Although you stated you are not interested in Huygens' principle, I want to add a note on this explanation. I will need in my answer the expression for the electric field of a radiating dipole
$$\boldsymbol{\rm E}\left(\boldsymbol{r},t\right)=-\frac{\omega^{2}\mu_{0}p_{0}}{4\pi}\sin\theta\frac{e^{i\omega\left(\frac{r}{c}-t\right)}}{r}\hat{\theta}$$
This expression assumes the dipole oscillates in the $\hat{z}$ direction. Now look for example at this picture
taken from here. This illustration seems to discard the polarization of the incoming wave (as you said), but if you think about it more - it turns out it doesn't. The radiation in the plane of incident is circular only for $s$-polarized light, since then each dipole is oscillating in the $z$ direction (in and out the page) and radiating a field given by
$$\boldsymbol{\rm E}\left(r,\theta=\frac{\pi}{2},\varphi,t\right)=-\frac{\omega^{2}\mu_{0}p_{0}}{4\pi}\frac{e^{i\omega\left(\frac{r}{c}-t\right)}}{r}\hat{z}$$
independent of $\varphi$ and $s$-polarized too. If, on the other hand, you want to treat $p$-polarized light, then each lattice point should radiate like in this image
taken from here, and it will definitely have other consequences from the $s$-polarized dipoles. A popular example is the existence of the Brewster's angle, which is the result of the dipole not radiating on its oscillations axis. Also, as before, you can see that the polarization of the far field radiation is parallel to the direction of the oscillations of the dipole. This means that the $p$-polarization is maintained.
You are right to question the assumption that the impulse given to the mirror is twice the incoming photon's momentum, but this is not anything to do with the mirror being a perfect reflector. This is an approximation. You are correct that if the reflected photon's momentum is equal to the incoming photon's momentum, then the mass of the mirror must be infinite (or else, the reflected photon's momentum must be lower). This is another way of saying that the mirror cannot move. The best way to explain is to just do a simple kinematic calculation.
Let's ignore the spring and do a simple elastic collision of mirror and photon. This is a non-relativistic calculation (nothing like Compton scattering), so let's use non-relativistic energy and momentum conservation.
Let's take the incoming photon's wavelength to be $\lambda$, the reflected photon's wavelength to be $-\lambda^\prime$ (negative since it is reflected in direction), and the mass of the mirror as $M$. After the photon reflection, suppose the mirror is imparted a velocity $v$. By momentum conservation,
$$
\frac{h}{\lambda} + \frac{h}{\lambda^\prime} = M v \:,
$$
and by energy conservation,
$$
\frac{hc}{\lambda} - \frac{hc}{\lambda^\prime} = \frac{1}{2}Mv^2\:.
$$
You can convince yourself that after eliminating $h/\lambda^\prime$ and some rearranging you will have an equation that is quadratic in $v$, whose formal solutions will be
$$
v = -c \pm c\sqrt{1+\delta} \:,\:\:\:\:\:\:\textrm{where}\:\:\:\delta = \frac{4 h}{M c \lambda} \: .
$$
I'm skipping a lot of trivial algebra (and the quadratic formula); you should be able to get the result above without too much trouble. We can immediately throw away the unphysical tachyonic solution, and, since $\delta \ll 1$, we can expand in $\delta$ to get
$$
\frac{v}{c} = \frac{1}{2}\delta + O(\delta^2) \:.
$$
Thus, we get
$$
Mv \approx \frac{2h}{\lambda} = 2 p \:,
$$
where we have ignored terms that are higher order in $h/\lambda$ (meaning higher order $\delta$ terms). Thus, the momentum of the mirror is approximately just twice the incoming photon's momentum. In other words, you can just approximate the kinematics of the system as if the reflected photon has the same momentum $p$ as the incoming photon, and that the mirror is therefore given a momentum $2p$ because the photon gets reflected (the photon's impulse must be $-2p$ in order to reverse direction, so the mirror's impulse must be $+2p$ in order to conserve momentum).
In reality, the photon will see some wavelength shift, but it will be small. The leading order term in the mirror's impulse will come from the photon's change of momentum due to reflection. Intuitively, this is because the rest mass energy of the mirror is much larger than the photon's energy. For the sake of intuition, you can pretend the photons are equivalent, here, to small particles of mass $m$, where $m$ is given by $m c^2 = hc/\lambda \ll M c^2$. Think of bouncing a marble on the ground, where the mass of the Earth is much larger than the mass of the marble: each individual photon's momentum will not change in magnitude by much since the mirror's mass is much higher, only its direction will change. This intuition is supported by the analysis above: we would expect our conclusions to break down when $\delta \sim 1$, or in other words, when $h/\lambda \sim Mc$ (ignoring trivial numerical factors).
As an aside, we can also approximate what the wavelength shift will be. The value of $v$ up to first order corrections will be
$$
\frac{v}{c} = \frac{1}{2}\delta - \frac{1}{8}\delta^2 + O(\delta^3) \: .
$$
Thus,
$$
Mv \approx 2 \frac{h}{\lambda} - \frac{2}{Mc} \left(\frac{h}{\lambda}\right)^2 \: .
$$
Placing this expression back into to conservation of momentum equation at the top, we will have
$$
\frac{h}{\lambda^\prime} - \frac{h}{\lambda} \approx - \frac{2}{Mc} \left(\frac{h}{\lambda}\right)^2 \: .
$$
So,
$$
\frac{\Delta p}{p} \approx - \frac{2p}{Mc} \:,
$$
where higher order corrections in $p$ will be suppressed by factors of $1/Mc$. In terms of $\lambda$, this shift will be, up to first order corrections,
$$
\frac{\Delta\lambda}{\lambda} \approx \frac{2h}{Mc\lambda} \:.
$$
So, if we take visible light (say, $\lambda = 5 \times 10^{-7} \,\textrm{m}$), and $M = 0.1 \,\textrm{kg}$, this proportional shift will be about
$$
\frac{\Delta\lambda}{\lambda} \approx 8 \times 10^{-35} \:,
$$
which is absolutely the textbook definition of negligible. Where the mirror picks up its detectable motion is in the sheer number of photons hitting it.
Best Answer
I think you are perhaps confusing reflection at a corner with a "corner reflector" (https://en.wikipedia.org/wiki/Corner_reflector). The latter is a type of mirror which reflects light back to where it came from, as shown in your diagrams. The reflection comes from the 3 perpendicular mirrors, rather than at the join between them.
There is no "specular" (mirror-like) reflection exactly in the corner formed by 3 mirrors, or along an edge formed by 2 mirrors. You will not see an image in such a corner or edge no matter what direction you look at it. Light which strikes at or very close to the corner or edge will be scattered or diffracted in a number of directions, because of the wave nature of light. Scattering depends on the details of how the corner is made, what imperfections it has, and the wavelength of the light. This is "diffuse" reflection, which occurs from microscopically irregular surfaces. If the join is very well made you may not notice by eye the actual corner or edge in the reflected image (unless you magnify it). Otherwise, if you do notice anything it will just be a thin dark line or spot, or (in magnification) a blurring of the image.
I think that your guess is wrong : a ray pointed into a corner will not be reflected directly back upon itself, even if the join is perfect on sub-wavelength scales. If that were the case you would not be able to see the edge or corner at all.