Lorentz Velocity Transformation: $$v_x = \frac{v_x' + u}{1+\frac{uv_x'}{c^2}}\tag1$$
$$v_x' = \frac{v_x - u}{1-\frac{uv_x}{c^2}}\tag2$$
The speed of spaceship and scoutship are given relative to Earth whereas the speed of the Robot Space Probe is given relative to the spaceship.
Assume we want to find the speed of Robot Space Probe relative to Earth I understand that the formula must use formula (1) where $v_x$ is the speed required. This is clearly obtainable from the fact that the Earth is at rest relative to the moving spaceship and hence this reduces to a situation similar to the usual "Observer - Train" situation. This is also recognizable from the derivation where we assumed that the reference frame S' moves at speed u relative to the reference frame S.
However, assume that we have to find the speed of the Scoutship relative to the Spaceship what Lorentz Transform Equation should we use? Here the case is kind of confusing because the two spaceships are moving what should we consider as our S frame and which one should we consider as S' frame?
I know that the math shows that taking the wrong reference frame gives us a value of magnitude greater than the speed of light and hence we would have to take the other Lorentz Equation. However, I need to learn the concept.
Images are taken from: University Physics with Modern Physics Sears and Zemansky 13th Edition.