Events and lorentz transform 
A and B both start at the origin and simultaneously head off in opposite directions at speed $3c/5$ with respect to the ground. A moves to the right, and B moves to the left. Consider a mark on the ground at $x = L$. As viewed in the ground frame, A and B are a distance $2L$ apart when A passes this mark. As viewed by A, how far away is B, when A coincides with the mark?

This question is essentially simple, and we can solve it going to A's frame, take the time the ground mark reaches A, and multiply this time by the relativistic velocity of B with respect to A.
However, I just tried to solve by another way, and I am confused why this way to solve is wrong. Consider the following:
Suppose two events on the ground reference system: Event $r$ is A reaches the ground mark, and event $s$ is "instantaneous measure of the position of B in spacetime when A reaches the ground mark".
In another words: Let $(t,x)$.
$$r:= (t,L)$$ $$s:= (t,-L)$$
$$\Delta t = 0, \Delta x = 2L$$
Now, LT to A frame, so that $\Delta x' = -\gamma 2L$
Now, this is obviously wrong. But I can't figure out why. Something tells me that the problem is with the $s$ event, but I don't really know what. I mean, there are a lot of problems where doors shut simultaneously, so the problem with the chosen events is not that the position of B is measured automatically when A reaches the mark. So, what is the problem?
 A: The problem is that your event $s$ refers to the point along B's trajectory that is simultaneous with the event $r$ in the ground frame. However, the question asks about "when A coincides with the mark, as viewed by A" (note that I rearranged the statement a little). The key idea is that "when A coincides with the mark" depends on which reference frame you're in, and the event you actually want to look at is the point along B's trajectory that is simultaneous with the event $r$ in A's frame.
A: You have forgotten to take into account the relativity of simultaneity. If you ask the question 'where is B now?', the answer will be dependent upon the location and frame of reference of the observer. Now is a relative concept. In a given reference frame, it is now at every point in space. However, the planes of simultaneity in frames that are moving relative to each other are tilted. If you are in one frame moving with respect to a second, your flat plane of simultaneity equates to a sloping plane of time in the second frame, the slope rising upwards in your direction of travel.
In the example you quoted there are three frames, and each will have its own plane of simultaneity tilted relative to the others. This means that where B is now from the perspective of the ground frame is not where B is now from the perspective of A.
