Let's replace the fiber optic cable with one laser source and one photodetector a distance $L=186,000$ mi apart in vacuum and at rest relative to each other. The laser source is pointed straight at the photodetector. Alice observes the laser source and the detector moving at constant velocity $v = 93,000$ mi/s $= c/2$ with respect to her inertial frame, in the positive $x$ direction. The source fires a pulse at the exact moment $t=0$ when it passes by Alice. So according to Alice, how long does it takes the light pulse to reach the detector? What if the source and the detector switch places? Or if the source-detector direction is perpendicular to the direction of motion?
Short answer: For $v=c/2$, Alice sees the light pulse hit the detector after $c\Delta t = L \sqrt{3}$ or $\Delta t = \sqrt{3} s$. When the source and the detector switch places, she observes the detection after $c\Delta t = L /\sqrt{3}$ or $\Delta t = 1/\sqrt{3} s$, while for a setup perpendicular to the direction of motion she sees $c\Delta t = 2L /\sqrt{3}$ or $\Delta t = 2/\sqrt{3} s$.
Longish details:
1. Forward propagating light pulse
Due to length contraction, at $t = 0$ Alice sees the detector at position $x = L/\gamma$, with $\gamma = (1-\beta^2)^{-1/2}$ the time dilation factor and $\beta = v/c$. So she finds that it travels according to
$$
x_{\text{detector}}(t) = \frac{L}{\gamma} + \beta ct
$$
She also sees the light pulse traveling at light speed, according to
$$
x_{\text{pulse}}(t) = ct
$$
The pulse reaches the detector when $x_{\text{pulse}}(\Delta t) = x_{\text{detector}}(\Delta t)$, meaning
$$
c\Delta t = \frac{L}{\gamma} + \beta c\Delta t \;\; \Rightarrow \;\;c\Delta t = \frac{L}{(1-\beta)\gamma} = L\sqrt{\frac{1+\beta}{1-\beta}}
$$
For $\beta = 1/2$ this gives
$$
c\Delta t = L \sqrt{3}
$$
Here's the confusing issue:
In the source-detector frame the corresponding duration is clearly $c\Delta t' = L$. But if time in the source-detector frame appears time-dilated to Alice, such that $c\Delta t' = c\Delta t/\gamma$, then how is it that for $c\Delta t'=L$ we find $c\Delta t = \frac{L}{(1-\beta)\gamma}$ instead of $c\Delta t = L\gamma$?
If we look carefully, Alice determines the pulse propagation time as the time during which she observes the detector moving from
$$
x_{\text{detector}} = \frac{L}{\gamma}\;\;\text{at}\;\;ct = 0
$$
to
$$
x_{\text{detector}} = \frac{L}{\gamma} + \frac{\beta L}{\gamma(1-\beta)} = \frac{L}{\gamma(1-\beta)}\;\;\text{at}\;\;ct = \frac{L}{\gamma(1-\beta)}
$$
In the source-detector frame, the end point indeed corresponds to coordinates
$$
x'_{\text{detector}}\Big|_{ct = \frac{L}{\gamma(1-\beta)}} = \gamma\left(\frac{L}{\gamma(1-\beta)} - \beta \frac{L}{\gamma(1-\beta)} \right) = L\\
ct'_{\text{detector}}\Big|_{ct = \frac{L}{\gamma(1-\beta)}} = \gamma\left(\frac{L}{\gamma(1-\beta)} - \beta \frac{L}{\gamma(1-\beta)} \right) = L
$$
as we'd expect for a light pulse propagating across distance $L$ starting at $ct=0$, but the start point has coordinates
$$
\begin{eqnarray}
x'_{\text{detector}}\Big|_{ct=0} &=& \gamma\left(\frac{L}{\gamma} - \beta \cdot 0\right) = L \\
ct'_{\text{detector}}\Big|_{ct=0} &=& \gamma(0 - \beta \frac{L}{\gamma}) = -\beta L\;<\;0 \;\;(!!)
\end{eqnarray}
$$
According to source-detector time, Alice starts monitoring the detector at a time $ct'= -\beta L$, before the pulse was ever emitted! The detector's proper time duration between Alice's start and end of observation events is then
$$
c\bar{\Delta t'} = L - (-\beta L) = (1+\beta) L \;>\;L
$$
which is indeed just the time dilated propagation time observed by Alice, as it has to be:
$$
c\bar{\Delta t'} = \frac{c\Delta t}{\gamma} = \frac{L}{\gamma^2(1-\beta)} = L(1+\beta)
$$
On the other hand, the start of observation event in the source-detector frame is at $x'=L$ and $ct'= 0$, which to Alice means
$$
x = \gamma(L + \beta \cdot 0) = \gamma L\\
ct = \gamma(0 +\beta L) = \beta\gamma L
$$
So for Alice the time interval from this event until the light pulse reaches the detector is
$$
c\bar{\Delta t} = \frac{L}{\gamma(1-\beta)} - \beta\gamma L = \frac{\gamma L}{(1-\beta)}\left(\frac{1}{\gamma^2} - \beta(1-\beta) \right) = \gamma L
$$
as expected from time dilation for $c\Delta t' = L$.
Note that Alice sees the light pulse closing in on the detector at an apparent relative velocity
$$
\Delta v = \frac{d}{dt}\left[x_{\text{detector}}(t) - x_{\text{pulse}}(t) \right] = -(1-\beta)c = v-c
$$
2. Backward propagating light-pulse
Suppose now the source and detector switch places. The source fires at $t=0$ in Alice's time, when it passes the $L/\gamma$ mark on her $x$-axis, and in the negative $x$ direction. At the same moment Alice sees the detector at $x=0$. She observes the light pulse propagating according to
$$
x_{\text{pulse}}(t) = \frac{L}{\gamma} - ct
$$
and the detector according to
$$
x_{\text{detector}}(t) = \beta ct
$$
This means, as before, that the light pulse hits the detector when $x_{\text{pulse}}(\Delta t) = x_{\text{detector}}(\Delta t)$ or
$$
\frac{L}{\gamma} - c\Delta t = \beta c\Delta t \;\;\Rightarrow \;\; c\Delta t = \frac{L}{\gamma (1 + \beta)}
$$
For $\beta = 1/2$, $\gamma = 2/\sqrt{3}$ we now have
$$
c\Delta t = \frac{L}{\sqrt{3}}
$$
Again, this is different from $c\Delta t = L\gamma$ as would be expected from simple time dilation, and for a similar reason.
In the source-detector frame the source fires from location $x'=L$, but at time $ct' = -\beta L$, and the pulse propagates until it meets the detector at
$$
\begin{eqnarray}
x' &=& \gamma\left(\frac{\beta L}{\gamma(1+\beta)} - \beta \frac{L}{\gamma (1 + \beta)}\right) = 0\\
ct' &=& \gamma\left(\frac{L}{\gamma (1 + \beta)} - \beta\frac{\beta L}{\gamma(1+\beta)}\right) = (1-\beta) L
\end{eqnarray}
$$
so the proper propagation time is indeed
$$
c\Delta t' = (1-\beta)L - (-\beta L) = L
$$
But Alice's start event corresponds to the detector being at $x=0$ for $t=0$, which is just $x'=0$ for $t'=0$ in the other frame. In the latter her observation time $c\Delta t = \frac{L}{\gamma (1 + \beta)}$ corresponds to
$$
c\bar{\Delta t'} = (1-\beta)L - 0 = (1-\beta)L
$$
which is, of course, the expected time dilated value, $c\bar{\Delta t'} = c\Delta t /\gamma$.
Note that now Alice observes the light pulse closing in on the detector at apparent relative velocity
$$
\Delta v = \frac{d}{dt}\left[x_{\text{detector}}(t) - x_{\text{pulse}}(t) \right] = \beta c - (- c) = v+c \;>\; c
$$
However, this is not superluminal propagation! It is just the rate at which the distance between two simultaneous events in Alice's frame (observation of detector and observation of light pulse) changes in time. There are no objects moving at velocity $v+c$.
3. Light pulse propagating perpendicular to the direction of motion
Finally, when the source and the detector are arranged perpendicular to Alice's direction of motion, we have a variant of the light clock setup. In this case Alice observes the light pulse going at light speed in a direction tilted relative to the direction of motion of the source at a slope $\frac{1}{\beta\gamma}$ and hitting the detector after a time $c\Delta t = \gamma L$. This answer explains in more detail why this happens.
If you prefer to consider a fiber optic cable where light propagates at $v \sim 2c/3 < c$, as suggested in one of the comments, simply use the velocity addition formula to obtain the velocity of a pulse as seen by Alice, then apply the same reasoning.
