Throughout this answer we'll use "feel" or "felt" to mean the force or acceleration that is perceived by the pilot and transmitted through the aircraft structure as stresses or strains. In a "weightless" situation, the net force or acceleration is not zero, but the "felt" force or acceleration is zero.
It is valid to argue that the force or acceleration "felt" by the pilot, and registered by the instruments, in flight is simply the net aerodynamic force or acceleration generated by the aircraft. To get the actual net force acting on the aircraft, we have to add gravity to the aerodynamic forces. In other words, the "felt" force or acceleration is the net force or acceleration acting on the aircraft, minus the force or acceleration due to gravity.
1) Straight and level upright flight-- net acceleration zero, net force zero, "felt" acceleration 1-G in the upward direction, force "felt" by the pilot is simply equal to his weight, acting in the upward direction. This force is the aerodynamic lift force generated by the wings, transferred through the aircraft structure to the bottom of the seat and then to the pilot's body.
2) Aircraft inverted at the top of a loop-- again the acceleration and force "felt" by the pilot are simply the result of the net aerodynamic force generated by the aircraft. If the wings are generating positive lift, the pilot and G-meter will feel positive G's, and if the wings are generating negative lift, the pilot and G-meter will feel negative G's. If the wings are generating zero lift, the pilot and G-meter will feel zero G's (weightlessness).
3) Steady-state coordinated turn-- the net aerodynamic force generated by the aircraft is simply the wing's lift vector, which contains no lateral (sideways) component in the aircraft's reference frame. The slip-skid ball is centered. Meanwhile the net force including gravity is the purely horizontal force vector-- the centripetal force vector-- that drives the turn. This force vector does have a lateral (sideways) component in the aircraft's reference frame, yet the slip-skid ball is centered.
4) Turn with some sideslip-- the nose of the aircraft is allowed to point toward the outside of the turn. The airflow impacts the side of the fuselage and generates an aerodynamic sideforce. Now the net aerodynamic force generated by the aircraft does include a lateral (sideways) component, so the slip-skid ball rides toward the inside of the turn, and the pilot's body tends to lean toward the wall of the cockpit that is on the inside of the turn.
Yet if we take our airplane and bolt it to a roller coaster and run it through loops, flat turns, etc, the pilot still "feels" apparent forces, and the G-meter and the inclinometer still give indications. What are causing these apparent forces? (Assume for simplicity that the roller coaster is in a vacuum.)
We can say that these apparent forces are entirely due to the forces exerted by the rails of the roller coaster upon the wheels of the roller coaster, with gravity playing no role. This is analogous to saying that the forces "felt" in flight are entirely due to the aerodynamic forces created by the aircraft, with gravity playing no direct role.
Yet we can also say that these apparent forces are entirely due to the combined effects of gravity, and of "centrifugal force". Obviously, if we don't start a loop in the roller-coaster with enough speed to maintain positive G's all the way around, we'll hang upside-down in the seatbelts as we go over the top.
The key to reconciling these viewpoints is to understand exactly what "centrifugal force" is. It is the apparent force created by a curvature in a vehicle's trajectory. It is not a real force. It is basically just the mirror image of the real forces that are acting to curve a vehicle's trajectory. For (almost) every centripetal force acting to drive a curvature in a vehicle's trajectory, we can say the occupants "feel" an apparent "centrifugal force" acting in the other direction. When we do a high-speed turn to the left in a car on flat ground, the tires generate a real force toward the left, and we say we perceive a centrifugal force to the right, and are thrown against the right wall of the inside of the car. If we want to understand which direction a slip-skid ball would deflect, we could consider the real centripetal force from the tires, or the apparent "centrifugal force" generated by the resulting turn rate, but not both. If we consider both forces, we end up with zero net force-- they would cancel each other out-- predicting that the slip-skid ball would not deflect at all. This would be wrong. We can't mix and match apparent "centrifugal force" with real centripetal force. We have to use one or the other.
In this context, we're not using "apparent" to mean "the felt component of". We're just using "apparent" to remind ourselves that "centrifugal force" doesn't actually exist.
We stated that "For (almost) every centripetal force acting to drive a curvature, we can say the occupants 'feel' a "centrifugal force" acting in the other direction". Why "almost"? Because we can't "feel" any centrifugal force that is caused by a curvature in a vehicle's trajectory that is due to gravity. (Consider the forces felt by an astronaut in orbit.)
Why is gravity such a special force? Forces from wings and forces from wheels must be transmitted through the aircraft structure and to the exterior of the pilot's body. As a result, stresses and strains are created in the aircraft structure and in the pilot's body. In contrast, gravity exerts an equal acceleration per unit mass on every molecule of the aircraft structure, and on every molecule of the pilot's body, creating no stresses or strains, i.e. no displacement of one part of the structure or body relative to another part. Likewise, unless we consider the outer housing of the slip-skid ball or the G-meter to be rigidly fixed in space, gravity has no tendency to displace any element of either of those instruments in relation to any other element. For example, gravity has no tendency to pull the inclinometer ball to one side within the glass tube, even if the tube is tilted relative to the earth. (We're assuming that our instrument is much smaller than planetary in scale, so that tidal effects may safely be ignored!)
We've defined "centrifugal force" as an apparent force that is the mirror image of the net centripetal force acting to drive a curve in a vehicle's trajectory. If the force of gravity is not "felt", then what is the "felt" component of that apparent "centrifugal" force?
Since gravity is already "built in" to the net centripetal force determining a vehicle's trajectory, it follows that an apparent upward force opposite to gravity is already "built in" to the apparent "centrifugal" force that is paired with that centripetal force. This means that the "felt" centrifugal force is equal to the total centrifugal force plus gravity. Just as the "felt" centripetal force is equal to the total centripetal force minus gravity.
(Note-- it's really only the component of gravity that acts perpendicular to the flight path that is "built in" to the centripetal force vector. For simplicity, in the discussion of "centripetal force" in this answer, we'll assume that we're looking at the vehicle's trajectory at a point where the trajectory is exactly horizontal rather than rising or falling, and thus the full weight vector acts perpendicular to the flight path. Alternatively, we could think in terms of a three-dimensional vector sum rather than a two-dimensional one.)
The key point here is not to argue about whether "centrifugal force" actually exists or not. Rather the key here is to recognize that we're starting from a situation where the exact trajectory is known, and then we're calculating the centripetal (or centrifugal) force that is involved in creating this trajectory, and then we're subtracting away (or adding) gravity to get the "felt" component of that centripetal (or centrifugal) force.
In essence, "centripetal force" is just another way to say "net force", except that we're discarding components that act tangential rather than perpendicular to the flight path. In the context of flight, the aerodynamic forces and the force of gravity are both built in to the concept of "centripetal force". "Centrifugal force" is an apparent force or pseudoforce that is equal and opposite to the real "centripetal force".
In the case of a roller coaster, the "felt" component of the centripetal force would be the force exerted by the tracks on the wheels, and the "felt" component of the centrifugal force would be equal and opposite. In the case of an aircraft, the "felt" component of the centripetal force would be the aerodynamic force created by the aircraft, and the "felt" component of the centrifugal force would be equal and opposite. Whether we want to say that the pilot, G-meter, and inclinometer are responding to the "felt" component of the centripetal force or the "felt" component of the centrifugal force is really just a matter of convention, but only the former actually exists.
So all these three things are true:
1) "Felt" force = aerodynamic force = net force minus gravity
2) "Felt" force = net centripetal force minus gravity
3) "Felt" loading (the sensation of apparent weight due to being accelerated by a force) = ( net "centrifugal" force plus gravity ) / mass
We're taking the convention here that the felt "loading" acts in the opposite direction to the felt "force" or felt "acceleration", just as the apparent "centrifugal" force acts in the opposite direction to the actual centripetal force.
Whether we want to work in terms of "centripetal" or "centrifugal" forces is really just a matter of convention. Only the "centripetal" forces are actually real. To get the "felt" component, we subtract gravity in the former case, and we add gravity in the latter case. Because of the way that gravity can't actually be "felt", yet it is already "built in" to the centripetal and "centrifugal" forces.
Note also that by specifying "centripetal" (or "centrifugal"), we're saying that we're limiting the force and acceleration components of interest to those acting perpendicular (orthogonal) to the flight path, and disregarding those acting parallel (tangent) to the flight path. The first equation in the list of three above does not have that constraint.
Often one sees diagrams in flight training manuals that try treat "centrifugal force" as just one component out of several forces acting to curve the flight path. Many diagrams have been published, including by the FAA, that illustrate the deflection of the slip-skid ball in turning flight as being due to a force balance between "centrifugal force", and "gravity", and lift, or between "centrifugal force" and lift. (Oddly, the real aerodynamic sideforce caused by the fuselage flying sideways through the air is invariably omitted.) At first this seems intuitively right-- if you are bringing gravity into the picture rather than just purely considering centrifugal force alone, then why not bring lift into the picture too? But this approach just doesn't work. The effect of the lift vector is already included in the centrifugal force vector. To explain the forces "felt" in a coordinated or uncoordinated turn, we need to work in terms of the real aerodynamic forces, or we need to work in terms of the balance between "centrifugal force" and gravity, which is essentially the same as working in terms of the balance between centripetal force and gravity. A diagram combining gravity, centrifugal force, and lift just can't hold water.
If the flight path is fully constrained, then the "centrifugal force plus gravity" (or "centripetal force minus gravity") approach works well. For example, if the flight path is known to be linear, then centrifugal force is known to be zero, and the slip-skid ball works as a bubble level, telling us which way is down.
But in many situations, the flight path is not constrained to be linear. Also, unlike the case of the roller coaster, the turn radius is free to vary as the lift force is varied, and the flight path is free to curve up and down in the vertical plane as well. That is why it is so useful to take the viewpoint the forces "felt" in flight are simply the real aerodynamic forces generated by the aircraft, with gravity and "centrifugal force" playing no direct role.
Consider for example this related ASE answer about why a glider experienced a negative G-load at the top of the loop. If we imagine the loop to be constrained to follow a pre-defined track with as set radius like a roller coaster, we could imagine that the explanation is simply that we didn't start with enough speed to generate enough "centrifugal force" to counteract gravity and "make it all the way around" without hanging loose in the belts. But recognizing that the flight path is not constrained to follow a pre-defined track, it is more useful to recognize that when we "felt" a negative G-loading, the wing must have been flying at a negative angle-of-attack and generating negative lift. Which raises the question, "Why was the wing generating negative lift-- with the stick full aft? Why didn't the angle-of-attack stay positive, as the pilot was intending to command, even if the resulting trajectory did not result in a perfectly round loop?" The negative-lift situation is in fact related to the low airspeed at the top of the loop, but for reasons that are much more subtle than the fixed-track roller-coaster analogy would suggest.
The idea that a coordinated turn features a "balance" between centrifugal force and gravity may be technically true, but it has no explanatory power, especially in the context of a flight path that is free to vary in three dimensions as lift and other variables are varied, rather than being constrained to run along a pre-existing track as in the case of a roller coaster. It doesn't reveal that the key factor in a "coordinated" turn is simply that the fuselage is streamlined to the airflow, so that the aircraft is generating no aerodynamic sideforce and thus has no aerodynamic force component in the lateral (sideways) direction in the aircraft's reference frame. It can lead one to believe that in a sideslip, an abnormally low turn rate is somehow causing the aircraft to slide sideways through the air, rather than understanding that the aircraft is being forced or allowed to fly sideways through the air which is creating aerodynamic sideforce (as well as drag), which is in causing a reduction in the turn rate. It can lead one to believe that a pilot should somehow be making adjustments in angle-of-attack to get the "right" lift force for the airspeed and bank angle, or else the aircraft will go sliding off toward the inside of the turn. (Footnote to be added.)
In short, the "centrifugal force plus gravity" approach is often not the most useful way to look at the problem when the flight path is free to vary rather than constrained to be some particular shape like a straight line, or a perfectly circular loop of pre-defined radius, or a round horizontal circle with a pre-defined turn radius and bank angle or airspeed. Unless we are back-engineering the "felt" forces (i.e. the aerodynamic forces) from a saved track log that was recorded with tremendous precision, the "centrifugal force plus gravity" approach doesn't have much application. To know what the "centrifugal" force is, we have to know what the trajectory is, and to know what the trajectory is, we usually have to know the forces acting on the aircraft, including the aerodynamic forces, in which case we already know the answer to question of what will be the result of "centrifugal force plus gravity".
The "centrifugal force plus gravity" approach is generally better suited to understanding the forces felt in race cars travelling along tracks of fixed bank angles and fixed turn radii, or the angle taken by a rope of fixed length tied to bucket full of water swung round in a circle, than to the forces "felt" in an airplane whose path is free to vary in three dimensions. But when applied properly, it is not an invalid way to look at the mechanics of flight.