I agree that the person on the embankment will say that the person on the train shouldn't see them as simultaneous
Well, then the person on the train shouldn't see them as simultaneous. Some things change between reference frames, but conclusions of the form "in frame $S$, an observer will see..." do not change, since the statement itself specifies which frame you have to be in to understand what it is saying.
The observer on the embankment could easily see the train observer intercept the forward flash before the rear flash. (Of course, the embankment observer couldn't do this in real time; one has to wait until after one's hypothetical grid of rulers and clocks reports back what happened when and where.) One nice thing about SR is that time-ordering is invariant. That is, two events $A$ and $B$ can have one of three relations to one another: $A$ is in $B$'s past light cone (and $B$ is in $A$'s future), the reverse of that statement, or $A$ and $B$ are spacelike separated. Whichever one of these holds will hold for all observers.
So we know, just from the embankment analysis, that "in the frame of the train, the forward flash reaches the observer's eyes first," and this statement is always true for anyone who speaks it in its entirety.
What about the train observer? Indeed, as you say,
the flashes occurred at the same distance from each of them, the speed of light is constant in both frames, and either one can claim to be at rest
Suppose two people, $C$ and $D$, stand equal distances from you and are known to pitch balls at exactly the same speed. With everyone standing at rest, $C$ and $D$ each toss you a ball. You get the ball from $C$ before the one from $D$. This is not a logical inconsistency. It simply means $C$ threw a ball before $D$ in your reference frame. That is, the person on the train, operating under the SR assumption of "the speed of light is constant," and using the data (retroactively obtained from a ruler-clock grid, or maybe obtained in real time based on brightnesses) that the flashes were equidistant, must conclude that the forward flash went off first.
@JohnDuffield is indeed correct. Let me try to quantitatively show the anisotropy of the speed of light for an observer on the rotating platform. I think the following reasoning is to be credited to Langevin (I don't remember the reference, sorry, and anyway it would be in French!).
So let's denote by $\omega$ the angular speed of the platform with respect to the still observer. Clocks and rods on the rotating platform will be affected by their motion with respect to the still observer. What order in $\omega$ shall we expect? If there is a term proportional to $\omega$, then the clocks and rods will behave differently depending on whether the platform rotate in one or the other direction. This is silly, as this amount to simply having the still observer look from the top instead of from the bottom for example. So the first term has to be proportional to $\omega^2$. But then, that means that if we limit ourselves to an approximation at order $\omega$, we can completely neglect the fact that clocks and rods on the rotating platform will not measure time and lengths as for the still observer. What that means is that we can use a good old Galilean transform!
So let's denote by $(x, y, z, t)$ and $(x',y',z',t')$ the spacetime coordinates on the platform and of the still observer respectively, then
$$\begin{align}
x'&=x\cos\omega t-y\sin\omega t\\
y'&=x\sin\omega t + y\cos\omega t\\
z'&=z\\
t'&=t
\end{align}$$
In differential form, this reads
$$\begin{align}
dx'&=dx\cos\omega t-dy\sin\omega t - \omega (x\sin\omega t+y\cos\omega t)dt\\
dy'&=dx\sin\omega t + dy\cos\omega t + \omega(x\cos\omega t-y\sin\omega t)dt\\
dz'&=dz\\
dt'&=dt
\end{align}$$
Then we look at the spacetime interval $ds$. The still observer being inertial,
$$ds^2 = dt'^2-dx'^2-dy'^2-dz'^2.$$
I have used units where the speed of light $c=1$. Then substituting the above change of coordinates, and a bit of trigonometry, gives
$$ds^2 = dt^2 -2\omega(xdy-ydx)dt\underbrace{-dx^2-dy^2-dz^2}_{-dl^2}.$$
The propagation of a light signal corresponds to $ds^2=0$. The presence of cross terms $dxdt$ and $dydt$ results in an anisotropic speed of light. Let's investigate that precisely.
We note that $dA=\frac{1}{2}(xdy-ydx)$ is the area of the infinitesimal triangle whose vertices are the origin of the coordinates, the point $(x,y,z)$ and the point $(x+dx, y+dy, z+dz)$, that is to say the area swept by the vector from the origin to the position of the light signal during the duration $dt$. This is an area with a sign: positive if the sweeping is anti-clockwise and negative otherwise. So we get
$$dt^2 - 4\omega dAdt -dl^2=0.$$
We can solve this 2nd order equation in $dt$:
$$dt = 2\omega dA + \sqrt{dl^2+4(\omega dA)^2}.$$
But since we neglect terms of order $\omega^2$,
$$dt = dl + 2\omega dA,$$
i.e.
$$\frac{dl}{dt} = 1 - 2\omega \frac{dA}{dt}.$$
This is the speed of light for an observer on the platform: not only it is not equal to 1 but it depends on the direction of propagation since it depends on the sign of $dA/dt$.
Best Answer
You are mistaken. Yes, nothing can move faster than light in a sense that physical motion is fundamentally bounded from above by $c$. But an abstract mathematical quantity, which is the distance between the light beam and the back wall of the wagon, can decrease with a slope higher than $c$. There is nothing in Special Relativity that forbids it from happening.
On the contrary, we know that the light beam is traveling to the left exactly with the speed of light $c$, because this is physical motion we are talking about. And since the wall is moving to the right, the distance decreases faster than $c$. Again, this is completely normal.
Consider another thought experiment: two particles flying in the opposite directions with almost the speed of light each. The distance between them obviously increases with speed $2c$. There is nothing wrong with this! However, in the rest frame of one of the particles, the other one travels not with $2c$, but approximately with $c$ because it is the way the relativistic velocity addition functions.