Solution

The key idea in this problem is that light takes a finite, non-zero time to travel from one place to another.

First, the signals were sent by the spaceship when the spaceship was at distances $x_1$ and $l-x_1$ from the stations. Henceforth, we will work in the frame where the stations are at rest for this solution. Also, the signal traveling to the left is signal $1$ and is reflected by the first station, and the signal traveling to the right is signal $2$ and is reflected by station $2$.

To calculate $x_1$ we need to use the idea that the signals, after being reflected, reach the ship at the midpoint between the stations; otherwise the signals would not reach the ship at the same time.

The time taken by ray $1$ to reach the spaceship at the midpoint is:

\begin{equation} t_1 = \frac{x_1 + l/2}{c} \end{equation}

On the other hand, the spaceship will move a distance $l/2 - x_1 = v t_1$ in that time. Substituting the time into this equation yields:

\begin{equation} \frac{x_1 + l/2}{c} = \frac{l/2 - x_1}{v} \;\rightarrow\; x_1 = \frac{l}{2} \frac{1-v/c}{1+v/c} \end{equation}

Knowing this, we can find the time difference between signals at one of the stations. The difference $\Delta t$ arises because ray $1$ takes a time
$t_{11} = \frac{x_1}{c} = \frac{l}{2c} \frac{1-v/c}{1+v/c}$ to travel from the emission point to the first station. On
the other hand, ray $2$ takes a time $t_{22} = \frac{l-x_1}{c} = \frac{l}{2c} \frac{1+3v/c}{1+v/c}$ to reach the second station, and a time $t_{21} = \frac{l}{c}$
to travel from the first station to the second station (or vice versa). Thus, the difference between the reflection times as seen from station $1$ is:

\begin{equation} \Delta t = t_{22} + t_{21} - t_{11} = \frac{l}{c} + \frac{2l (v/c)}{1+v/c} = \frac{l}{c} \frac{1 + 3v/c}{1+v/c} \end{equation}

Finally, when the spaceship sends the signals, it is at a distance $x_1$ from station $1$. The stations will observe the spaceship's position when the signals reach each station. Note that the signals take time to travel from the emission point to the stations. When signal $1$ reaches station $1$, the spaceship will be at a distance:

\begin{equation} x'_1 = x_1 + v t_{11} = \frac{l}{2} \left(1-\frac{v}{c}\right) \end{equation}

And when ray $2$ reaches station $2$, the spaceship will be at:

\begin{equation} x'_2 = l - x_1 - v t_{22} = (l - x_1)\left(1-\frac{v}{c}\right) = \frac{l}{2} \frac{1+3v/c}{1+v/c} \left(1-\frac{v}{c}\right) \end{equation}