What is the relative acceleration of a projectile fired at in a low gravity vacuum? This question is to end an old argument.
Given a (space) vehicle with an acceleration of X, and projectile with an acceleration of 2X (rocket, not bullet), what would the relative acceleration of the projectile be if fired from the vehicle along its axis of motion?
My assertion is that, relative to the vehicle, the rocket would accelerate at 2X, and to an outside observer, the rocket would accelerate at 3X (the acceleration of the launch platform and the projectile itself).
My friend’s assertion is that relative to the vehicle, the rocket would accelerate at X, and an outside observer would see the rocket accelerating at 2X.
 A: Your friend is correct.
The acceleration of the projectile is determined by the thrust its rocket motor can produce. If the acceleration of the projectile is $2g$ then the thrust of its motor would be $2mg$.
But launching the projectile from your space vehicle can't increase it's thrust. You can increase its initial velocity, but once the projectile has left the rocket all it has to push it is its motor. The thrust from the motor is $2mg$ whether it was launched from the rocket or from the ground, so its acceleration is $2g$ whether it was launched from the rocket or from the ground. Since the space vehicle accelerates at $1g$ their relative acceleration will be $1g$.
A: To make this easy, I will assume that all speeds are small compared to the speed of light $c$.
The motion is in one dimension, so the trajectory of the space vehicle will be the following:
$$x(t) = x_0 + v_0 t + \frac12 a t^2$$
We can easily set $x_0 = 0$ if the put the origin of the coordinate system at the start point where the space vehicle started. If the space vehicle started at rest at $t = 0$, then $v_0 = 0$ as well. So this makes it just:
$$ x(t) = \frac 12 at^2.$$
The rocket that is launched from the space vehicle at time $t_1$ will have the same position and velocity as the vehicle at launch. From that time on, it will accelerate with twice the acceleration, $2a$.
To make it simpler, assume that the rocket is launched directly at $t =0$, such that all the $x_0$ and $v_0$ terms are zero for the rocket as well. The rocket will have acceleration $2a$, so its trajectory is
$$ y(t) = \frac12 2 a t^2.$$
Since you have a space vehicle, it has to accelerate with some sort of repulsion mechanism, usually burning fuel (combustion engine) or acceleration ions (ion drive). The engine repulses itself off its fuel. The engine in the rocket does the same, it repulses itself off its own fuel, not off the space vehicle! Therefore, the relative acceleration is:
$$ \frac{\mathrm d^2}{\mathrm d t^2} [y(t) - x(t)] = 2a - a = a.$$
So from the space vehicle, the rocket is only accelerating with $a$ away from the space vehicle. An outside observer who is resting in the chose frame of reference will observe the accelerations to be $a$ for the space vehicle and $2a$ for the rocket.
If you change the problem a little bit, your way of viewing it would be correct. Say that the space vehicle had a long threaded rod that extends to the front of the space vehicle. If the rocket would propel itself against that rod (as opposed to propel itself against its fuel) the acceleration of both would add up, since then the position would be calculated like this:
$$z(t) = \frac12 2a t^2 + x(y) = \frac12 2a t^2 + \frac12 a t^2 = \frac12 3a t^2.$$
The first term is the acceleration of the projectile, the second is the one of the space vehicle. They add up and give you an acceleration of $3a$.
I think that your whole disagreement pivots around the understanding of propulsion in free space. See Newton's third axiom and the way rocket engines work.
