There are really two different questions here. You asked "What is the difference between centripetal and centrifugal force", but there's also an implication that you'd like to know what is the best answer to the homework problem. This answer will try to address both questions.
We could say that in a constant-speed coordinated turn, the real aerodynamic force is equal to the wing's lift vector, which if we desire we can break down into centripetal and vertical components. There is also a downward force component due to gravity. The net force components including gravity are in balance in the vertical dimension, but NOT in the horizontal dimension-- otherwise there would be no turn.
The forces the pilot "feels" are only the real aerodynamic forces, not gravity -- or perhaps it is more descriptive to say that the pilot "feels" an "apparent force" that is equal in magnitude and opposite in direction to the net aerodynamic force. So as the net aerodynamic force is transferred through the aircraft structure to the pilot's seat to the pilot's body, the pilot "feels" an "apparent force" pulling him down into his seat, which if we desire we can break down into centrifugal and downward components. The root cause of this "felt" or "apparent" force is the acceleration acting on the pilot's body-- excluding the component of acceleration due to gravity. Note that gravity does not actually "cause" the downward component of this apparent force -- if gravity were to instantly disappear the trajectory of the plane and pilot would both instantly change as the flight path curved upward into a loop but the pilot would not feel any change in the "apparent force" pushing him into seat. The fundamental reason for this is that gravity "works from within" and exerts an equal acceleration on every molecule of the pilot's body and aircraft without causing any stresses or strains (ignoring tidal effects), so it isn't perceived as trying to squish the pilot's body down into the seat. (We'd have to modify this point of view if we wanted to adopt a reference frame centered on the aircraft, treating the aircraft as a stationary object or an object moving at a constant velocity-- this would not be a valid inertial reference frame and would not tell us anything about the forces that a pilot actually "feels" in flight.)
In a constant-speed coordinated turn, thrust and drag are equal, and the plane is not being allowed to fly sideways through the air, so the airflow is not striking the side of the fuselage and generating an aerodynamic sideforce. So the net aerodynamic force is simply equal to the wing's lift vector. This force is acting straight "upward" in the aircraft's reference frame-- i.e. in same plane as the vertical fin-- and in response, the "apparent force" that is "felt" by the pilot is pulling him straight "down" into his seat-- i.e. in the same plane as the vertical fin.
Now what happens when we increase the angle-of-attack, causing at least a temporary increase in lift?
Let's take the point of view of looking at the real forces, not the "apparent forces". If we increase the wing's angle-of-attack, we increase the lift force, including the vertical and horizontal (centripetal) components. But we don't change the direction of the lift force. Saying that we've caused a change in centripetal force is true, but incomplete, because we're not mentioning the change in vertical force.
Now let's take the point of view of looking at the "apparent forces", not the real forces. If we increase the wing's angle-of-attack, we increase the lift force, including the vertical and horizontal (centripetal) components. This means that there's an increase in the "apparent force" pulling the pilot down into his seat, including both the vertical and horizontal (centrifugal) components. Saying that we've caused a change in the apparent centrifugal force is true, but incomplete, because we're not mentioning the change in the apparent vertical force.
If we just note the increase in the "apparent" centrifugal force, we might think that when we increase the angle-of-attack and lift force, the slip-skid ball (and the pilot's body) will tend to deflect (lean) toward the outside wall of the cockpit. This is not the case -- even though some of the very faulty diagrams we see in pilot training manuals and FAA exam materials might lead us to think otherwise.
The choice of 2) or 3) to the original question depends on whether we are interested in real forces or "apparent forces". But neither is a complete answer because both ignore the vertical force components at play. (If the question were about an increase in bank angle rather than an increase in angle-of-attack, then it would be a different story.)
2) is really a better answer than 3) because the question just asks about forces, not "apparent forces".
In the specific context of a coordinated turn, we could say that "centripetal" force is one component of the actual net force at play, while "centrifugal" force is one of component of the perceived force or "apparent force" at play, which is equal and opposite to the real force.
But more generally, "centripetal" means acting toward the center of the curve defined by the curving flight path (horizontal turn, or loop, or whatever), while "centrifugal" means acting away from the center of the curve traced out by the curving flight path (horizontal turn, or loop, or whatever).
We could have a discussion about performing loops that would be very similar to the discussion above. Again, the actual net force would have a "centripetal" component-- as would the portion of the actual net force that is due to aerodynamic force-- while the "perceived" force or "apparent force" would have a "centrifugal" component.
Yet there are other cases where we can generate an actual aerodynamic centrifugal force component that reduces the total aerodynamic centripetal force that the aircraft is generating. So it's not as simple saying that "centripetal" always refers to real force and "centrifugal" always refers to "apparent force". Example-- starting with a coordinated turn-- now apply lots of outside (top-side) rudder-- nose yaws up/out, airflow strikes side of fuselage creating an aerodynamic force toward high wingtip-- this is a real force, and it has a centrifugal component, so net centripetal force is reduced and the turn rate slows. Note also that when we add this new aerodynamic force into the picture, the total aerodynamic force is no longer acting in the same plane as the wing's lift vector, i.e. no longer is aligned with the vertical fin. Therefore the slip-skid ball will ride off-center toward the low side of the cockpit, and the pilot's body will tend to lean in that direction as well.
And to add yet another twist, consider an aircraft doing multiple loops without stopping. What is happening at the bottom of each loop? Gravity is a contributing a real centrifugal force component that affects the rate and radius of curvature of the flight path, yet the only "apparent force" the pilot "feels" is the apparent "centrifugal force" component that is exactly equal and opposite to the aerodynamic centripetal force generated by the wing.
Likewise we can think of a situation where the NET aerodynamic force is centrifugal rather than centripetal in nature. Example-- aircraft is flying an arcing trajectory such as the well known "Vomit Comet" zero-gravity simulator. We'll focus on the instant at the top of the curving arc, where "centripetal" is the same as "earthward" and "centrifugal" is the same as "skyward". Net aerodynamic force is zero, net "apparent force" is zero, total net force including gravity is equal to the weight of the aircraft and contents acting in the downward (centripetal) direction, and net acceleration is 1-G downward. Now if we repeat the same maneuver but with the wing generating a very small amount of lift-- say 1/10 the total weight of aircraft and contents-- we'll get almost the same arc. The net aerodynamic force is 1/10 the weight of the aircraft and contents, acting in the skyward (centrifugal) direction, so the "apparent force" acting on the aircraft, or on any object in the aircraft, will be equal to 1/10 the weight of that object, acting in the earthward (centripetal) direction. In other words, we'll "feel" 1/10 "G" of acceleration toward the earth. Our G-meter will read positive 1/10 "G". But the total net force including gravity is 9/10 the the weight of the aircraft and contents, acting in the earthward (centripetal) direction.