Forces in a body falling with air resistance

I would like to know why when a body falls with air resistance, we have that the drag caused by the air is $$-kv$$ and the acceleration is $$-mg$$.

What I don't understand is how they both have the same sign $$F_{\text(gravity)}=-mg$$ and $$F_{\text(drag)}=-kv$$.

I understand that the drag is negative cause is opposed to the velocity. I understand that the $$-mg$$ part is negative cause the vertical axis is positive in the upward direction. But I don't understand how come that air resistance has the same direction of the gravitational force. If I fall, I should feel an acceleration going down, and hence there's a force poiting downwards, and I should also feel a force going upward, cause of the air that I go through.

You are right that for a falling object, the force of gravity is down while the drag force points up (in the direction opposite to the velocity). And I understand your confusion from the equations - why is the force of gravity $-mg$?
This is because we usually write $g$ as a positive number, without taking account of its direction (even though it is pointing down). But when we write the drag force as $-kv$, we imply that $v$ is a vector $\vec{v}$ which therefore has a direction as well as a magnitude - and the drag force, which is also a vector, must point in the opposite direction.
It would be more consistent if we wrote acceleration as a vector $\vec{g}$, in which case the minus sign would disappear. But that is not the convention...