Acceleration transformation in special relativity

Vertical motion of the projectile means that energy is lost or gained by the projectile.

Horizontal motion of the projectile means that the the projectile moves relative to the aforementioned energy. The projectile receives a horizontal impulse when it absorbs some energy that has some horizontal momentum.

This is a much better scenario, because gravity is not involved:

A vertical rocket is in horizontal motion, rocket motors are off. When rocket motors are turned on, the horizontal motion of the rocket is unchanged, but the horizontal motion of an astronaut that roller skates horizontally inside the rocket changes.

If by projectile motion, you mean falling only under action of gravity, the acceleration in the frame of the object is zero. It is in free fall, and any test particle in that frame has no acceleration.

But we can suppose a rocket making a curve by using its engines, so that the crew will feel an acceleration.

The components of its 3-acceleration vector for an external inertial observer is:

$$a_i = frac{1}{1-v^2}left(frac{v_imathbf {v.a}}{1-v^2} + frac{dv_i}{dt}right)$$

If the inertial frame is momentarily moving with the rocket, $mathbf v = 0$, and the first term of the parentheses vanishes. The components of the acceleration $a_i = frac{dv_i}{dt}$, are the same as that measured by the rocket.

If the inertial frame is momentarily moving transverse to the rocket, $mathbf {v.a} = 0$. The acceleration measured by the inertial frame has the same direction, but is different in modulus from that measured in the moving frame.

In any other situation, the components of the acceleration measured by the inertial frame are different of that measured by the accelerated frame.