In a real-life misshapen blob of metal, strictly speaking the cyclic boundary conditions cannot be applied, since the blob only has a trivial group of spatial symmetries. However, the blob is approximately invariant under lattice translations (with the only mismatch occurring with the extremely small number of atoms at the surface) so it is tempting to associate the much larger symmetry group of lattice translation operators with the system.
Doing so, one obtains a system which is considerably more mathematically convenient than the otherwise rigorously-correct picture of a misshapen blob of atoms. As such, the Born-von Karman boundary conditions can be considered an approximate model to the behavior of real materials. That said, it's a pretty good approximation for macroscopically-sized materials.
That said, I don't have an explicit proof of why this approximation doesn't cause the math to explode and give wrong answers, so someone better versed than me could probably elaborate on this.
You have correctly derived your result on $k_x$. The key here is that, once you satisfy the condition on $k_x$ being an integer multiple of $\frac{2\pi}L$, any choice of $A$ and $B$ will not disrupt your boundary conditions, that are automatically satisfied. $A$ and $B$ can be arbitrary, and you only have a condtion relating them (the normalization of the wavefunction), so you have some freedom here. You can build wavefunctions that are not proportional to each other.
This is because you're solving for a free particle in a $1D$ box, so any eigenvalue of the energy $E=\hbar^2 k^2/2m$ is doubly degenerate: particles with impulses $\hbar k$ and $-\hbar k$ have the same enrgy, so the states $|\hbar k\rangle$ and $|-\hbar k\rangle$ are both eigenvectors for the Hamiltonian with the same eigenvalue, and they are linearly independent. The eigenspace relative to a certain eigenvalue $E$ has dimension $2$, and you have the freedom to mix different eigenstates of the impulse.
It is common use (and a convenient choice) to describe this eigenspace through a basis in which another useful operator is diagonal. As an example, one could use a basis of eigenvectors of the impulse (like you are doing in your question),
$$
\langle x|\hbar k\rangle=\frac{1}{\sqrt L}e^{i kx},\quad\quad
\langle x|-\hbar k\rangle=\frac{1}{\sqrt L}e^{-i kx},
$$
or a basis of eigenvectors of the parity operator, transforming $x$ into $L-x$:
$$
\langle x|+\rangle=\frac{1}{\sqrt {2L}}\left(e^{i kx}+e^{-ikx}\right),\quad \langle x|-\rangle=\frac{1}{\sqrt {2L}}\left(e^{i kx}-e^{-ikx}\right).
$$
Any way you choose the basis eigenvectors $|1\rangle$ and $|2\rangle$ (assuming that they are orthonormal), you have that any eigenstate of that energy $E$ is given by
$$
|\Psi\rangle=\alpha|1\rangle+\beta|2\rangle,
$$
with the normalization condition $|\alpha|^2+|\beta|^2=1$.
Best Answer
1.
Yes, the chosen $L$ influences the momentum eigenstates which are obtained from periodic boundary conditions. However, these momentum eigenstates are not the real eigenstates of the electrons.
The point of the boundary conditions is not to work out the physical states of the electrons but to count how many states there are. For a large system, we make the continuum approximation, i.e. we assume that the spacing between consecutive momenta is so small that it is practically a continuous variable such that the dependence of the momentum on $L$ is no longer important. Furthermore, we are normally interested in quantities per unit volume so the factor $V=L^3$ drops out, which is why the results are independent of $L$.
If you could actually measure the momentum of a single electron, its state would collapse into one of the allowed momentum eigenstates. However, these would not be the ones derived from periodic boundary conditions but from the actual boundary conditions of the system. Actual boundary conditions are generally only taken into account for systems where quantum confinement is important e.g. quantum dots (nanometre sized semiconductors which confine the electrons in all three dimensions) that can be modelled as a (finite) potential well. The results definitely $do$ depend on $L$ in these cases.
2.
The quantization volume is arbitrary and nothing to do with the physical sides of the metal. You can choose it to correspond to the actual metal but it is not necessary. All that matters is that it is big enough to make the continuum approximation.
Periodic boundary conditions yield complex exponential functions which 'happen to be' momentum eigenfunctions. They are a mathematical trick to give convenient functions to work with. Stationary wave boundary conditions on the other hand yield sine functions which are not momentum eigenfunctions. The introduction of a fake confining potential to achieve this is just another mathematical trick to fix the form of the functions. There are definitely not physical boundaries and no one is claiming that the walls of the metal approximate them. Indeed, as I said, the box doesn't even have to correspond to the sides of the metal.
The key point is that (focusing on one dimension for simplicity) any function $f(x)$ can be expanded on the interval $0<x<L$ as a Fourier series, but this can be either a complex exponential series or a sine series (or even a cosine series). The different types of boundary conditions are just ways of 'deriving' these different Fourier basis functions, any of which works equally well. Thus your statement
is not really correct as all boundary conditions are basically just a trick to obtain a certain form of basis function. The reason periodic boundary conditions are more popular is that momentum eigenstates are more convenient, e.g. we consider things like an electron with momentum $\boldsymbol{p}$ scattering into momentum $\boldsymbol{q}$.
As I said before, the size doesn't matter as long as it is large enough that we can take the continuum approximation and divide by the volume. I don't know why you would schematize it as a bunch of cubes and it is definitely not related to the lattice model of crystals: these boundary condition considerations apply in other cases such as particle physics and modes of an electromagnetic field.
As a final note, in Sommerfeld theory, the states obtained from periodic boundary conditions (momentum eigenstates) confusingly happen to be the actual energy eigenstates of the free electrons (in the continuum approximation). In general, the Hamiltonian and momentum operator do not commute, so eigenstates of momentum are not automatically eigenstates of energy. This is the case in a real metal, where electrons interact with the lattice resulting in Bloch states, superpositions of momentum states.