1) "When a quantum wave function is in a potential well, what causes the
quantization? The finiteness of the well, or only the term with ℏ in
Schrödinger's equation?"
For the quantum finite potential well, the discrete possible values for $E_n \sim \hbar ^2 v_n$ where the $v_n$ are discrete solutions to non-trivial equations due to the boudary conditions (see the details in the Wikipedia reference above). You may see directly in the formula, that both the Schrodinger equation (so quantum mechanics and $\hbar$), and the boudary conditions are necessary to have discrete values for $E_n$
2) Is there an analogy between these two approaches? Is the Schrödinger
equation fundamentally due to a sort of boundary condition, which
gives its value to the Planck constant ℏ?
No, this is not due to boudary conditions.
The basis of quantum mechanics is that position and momentum are no more commutative quantities, but are linear operators (infinite matrices), such that,at same time, $[X^i,P_j]= \delta^i_j ~\hbar$.
Now, you may have different representations for these operators.
In the Schrodinger representation, we consider that these linear operators apply on vectors $|\psi(t)\rangle$ (called states). The probability amplitude $\psi(x,t)$ is the coordinate of the vector $|\psi(t)\rangle$ in the basis $|x\rangle$. In this representation, you have $X^i\psi(x,t) = x^i\psi(x,t), P_i\psi(x,t) = -i\hbar \frac{\partial}{\partial x^i}\psi(x,t)$ . This extends to energy too, with $E\psi(x,t) = i\hbar \frac{\partial}{\partial t}\psi(x,t)$. This last equality is coherent with the momentum operator definition if we look at the de Broglie waves
3) One can obtain an analog of Schrödinger's equation if space was
discrete. Is it possible to derive Schrödinger equation from such a
description of space and time?
In the reference you gave, there is no discrete space, and there is no discrete time, the $\psi_i(t)$ are only the coordinates of the vector $|\psi(t)\rangle$ in some basis $|i\rangle$
Quantisation does not imply discreteness. If a system has been quantised, we just mean we have taken the set of states, and replaced it by a vector space of states. In other words, one can add states in quantum mechanics, allowing a system to be in two states "at once".
Observable quantities become certain operators acting on this vector space of states.
As you can see, this doesn't have anything, on the face of it, to do with discretisation. It turns out, however, that a lot of the operators we are interested in have discrete eigenvalues, and this implies that the corresponding physical values are discrete. Position, however, has a continuous spectrum, as do many other quantum observables.
There are plenty of sources explaining exactly how one goes from classical sets of states and numerical observables to quantum states (vector spaces - Hilbert spaces in particular) and quantum observables (operators); I won't cover that. All I shall say is that quantisation is a big mathematical process replacing a load of classical things with quantum things, and this sometimes leads to certain physical quantities being discretised.
Best Answer
Position quantization in vacuum is forbidden by rotational, translational, and boost invariance. There is no rotationally invariant grid. On the other hand, if you have electrons in a periodic potential, the result in any one band is mathematically the theory of an electron on a discrete lattice. In this case, the position is quantized, so that the momentum is periodic with period p.
Fourier duality
The quasimomentum p in a crystal is defined as i times the log of the eigenvalue of the crystal translation operator acting eigenvector. Here I give the definition of crystal position operators and momentum operators, which are relevant in the tight-binding bands, and to describe the analog of the canonical commutation relations which these operators obey. These are the discrete space canonical commutation relations.
In 1d, consider a periodic potential of period 1, the translation by one unit on an energy eigenstate commutes with H, so it gives a phase, which you write as:
$$ e^{ip}$$
and for p in a Brillouin zone $-\pi<0<\pi$ this gives a unique phase. The p direction has become periodic with period $2\pi$. This means that any superposition of p waves is a periodic function in p-space.
The Fourier transform is a duality, and a periodic spatial coordinate leads to discrete p. In this case, the duality takes a periodic p to a discrete x. Define the dual position operator using eigenstates of position. The position eigenstate is defined as follows on an infinite lattice:
$$ |x=0\rangle = \int_0^{2\pi} |p\rangle $$
Where the sum is over the Brillouin zone, and the sum is over one band only. This state comes with a whole family of others, which are translated by the lattice symmetry:
$$ |x=n\rangle = e^{iPn} |x=0\rangle = \int_0^{2\pi} e^{inp} |p\rangle $$
These are the only superpositions which are periodic on P space. This allows one to define the X operator as;
$$ X = \sum_n n |x=n\rangle\langle x=n| $$
The X operator has discrete eigenvalues, it tells you which atom you are bound to. It only takes you inside one band, it doesn't have matrix elements from band to band.
The commutation relations for the quasiposition X and quasimomentum P is derived from the fact that integer translation of X is accomplished by P:
$$ X+ n = e^{-inP} X e^{inP}$$
This is the lattice analog of the canonical commutation relation. It isn't infinitesimal. If you make the translation increment infinitesimal, the lattice goes away and it becomes Heisenberg's relation.
If you start with a free particle, any free $|p\rangle$ state is also a quasimomentum p state, but for any given quasimomentum p, all the states
$$ |p + 2\pi k\rangle $$
have the same quasimomentum for any integer k. If you add a small periodic potential and do perturbation theory, these different k-states at a fixed quasimomentum mix with each other to produce the bands, and the energy eigenstates $|p,n\rangle$ are labelled by the quasimomentum and the band number n:
you define the discrete position states as above for each band
$$ |x,n\rangle = \int_0^{2\pi} e^{inp} |p,n\rangle $$
These give you the discrete position operator and the discrete band number operator.
$$ N |x,n\rangle = n |x,n\rangle $$
if you further make the crystal of finite size, by imposing periodic boundaries in x, the discrete X become periodic and p becomes the Fourier dual lattice, so that the number of lattice points in x and in p are equal, but the increments are reciprocal.
This is what finite volume discrete space QM looks like, and it does not allow canonical commutators, since these only emerge at small lattice spacing.