Additional response to edited question: object is dropped with an initial groundspeed of zero: the only forces present as object falls are gravity and aerodynamic drag. Drag is a function of airspeed, which has a vertical component as well as a horizontal component. If you use the airmass reference frame, the ONLY significance of wind is that 1) the initial condition is that airspeed equals wind speed, and 2) during the time the object is falling there will be some horizontal translation between the airmass reference frame and the ground reference frame, which will affect the final answer when converted back to the ground reference frame. See notes below, and links at end, about not being able to solve for vertical and horizontal accelerations independently of each other.
Read on for more (some content was originally aimed at a broader version of the question) :
Cutting to the chase, is the object being dropped from a freely drifting balloon, or from a tower, or from an airplane? I.e. is the horizontal airspeed initially zero, or not? Dropping from an airplane is more like dropping from a tower than dropping from a freely drifting balloon, because the initial horizontal component of airspeed is not zero.
Once an object has accelerated to freely move with the wind, the horizontal component of airspeed is zero, so the wind exerts no force on the object. This makes calculations easier. In theory, if the initial horizontal component of airspeed is non-zero, it will take a infinite amount of time for the horizontal component of airspeed to fall all the way to zero, just as it takes an infinite amount of time to fully achieve the vertical terminal velocity. In practice, after a few tens of seconds the horizontal airspeed will probably be very close to zero.
If the object is initially released with zero groundspeed, then the horizontal component of the airspeed vector is initially not zero, but rather is equal to the wind speed. It turns out that this has an effect on the object's vertical acceleration as well as the object's horizontal acceleration, even if the body is a perfect sphere! (See links below for more.)
If the object is dropped from an airplane, wind or no wind, the situation is essentially the same as if the object is dropped from a tower in wind. The key point is that the horizontal component of airspeed is not zero at the instant of release.
If horizontal as well as vertical airspeed, force, and acceleration components are present at the instant of release, then you need to take care to properly consider the true magnitude and direction of the net aerodynamic force acting on the object. You need to write your equations in terms of the airspeed vector, which also is equal to the ground speed vector (including the vertical component) plus the wind speed vector. You can't treat drag due to wind speed, drag due to horizontal ground speed, and drag due to vertical falling speed as independent entities, because drag is dependent on airspeed squared, and all these components of motion contribute to airspeed.
Ultimately it's up to you whether you want to work in terms of the airmass reference frame (which moves with the wind) or the ground reference frame, so long as your equations appropriately deal with the x, y, and z components of the airspeed vector. It will generally be much easier to work in terms of the airmass reference frame, and view the ground as moving carpet down below, like one of those airport people-mover beltways. Just bear in mind that IF the object is released with zero groundspeed, then the initial condition is that the horizontal airspeed equals the wind speed. IF the object is dropped from a freely drifting balloon, then the initial condition is that airspeed is zero. IF the object is dropped from an airplane, then the initial condition is that the dropped object's airspeed equals the airplane's airspeed. Regardless of which is true, once the appropriate initial condition is set, forget about the wind, except for checking at the end to see how much the ground moved relative to the airmass reference frame, during the time of fall.
In other words, if the object is released from an aircraft with a known airspeed vector, and the airmass (wind field) is uniform, then you can ignore the wind while you do your calculations in terms of airspeed, and then you can add on the "drift" relative to the ground due to the translation of the airmass reference frame relative to the ground during the time of fall.
It appears from your question that you hoped to be able to compute horizontal acceleration components independent of vertical acceleration components. In any case where the object's initial horizontal airspeed is not zero, that's not going to work. Just as the horizontal airspeed at any instant affects the rate of change of vertical airspeed, so too does the vertical airspeed at any instant affect the rate of change of horizontal airspeed. See links below for more.
On the hand, if you are just dropping objects from a freely drifting balloon in a uniform wind field, then you can ignore most of this answer. In that case, horizontal airspeed, force and acceleration are zero.
These links should be of some use to you: see all associated links as well: