= Solution
The <photon> leaves the front rocket at $(t,x)=(0,a+L)$ and moves toward decreasing $x$, so its laboratory path is $x_\gamma(t)=a+L-ct$. Intersect it with the rear hyperbola, keeping the future root. Squaring and cancelling $c^2t^2$ gives
$$
\boxed{t_R=\frac{L(2a+L)}{2c(a+L)},\qquad
x_R=\frac{(a+L)^2+a^2}{2(a+L)}.}
$$
The rear rocket's speed there is
$$
\boxed{\beta_R=\frac{v_R}{c}=\frac{ct_R}{x_R}
=\frac{(a+L)^2-a^2}{(a+L)^2+a^2}.}
$$
At emission the front rocket is instantaneously at rest in S, so the <photon> laboratory <frequency> is $\omega_H$. At reception the rear rocket moves in $+x$ while the <photon> travels in $-x$. The <Doppler factor> for this head-on reception is
$$
\frac{\omega_F}{\omega_H}=\gamma_R(1+\beta_R)
=\sqrt{\frac{1+\beta_R}{1-\beta_R}}
=\boxed{\frac{a+L}{a}.}
$$
The receiver sees a blueshift. It is essential to use the front velocity at emission and the rear velocity at reception; comparing the two velocities at one laboratory time would describe the wrong pair of measurements. The result also agrees with the <Rindler light-transfer frequency ratio> for static accelerated observers at radii $a$ and $a+L$. In the weak-acceleration limit $L/a\ll1$, the shift is $1+L/a$, and the arrival time approaches $L/c$, providing useful checks.
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