Here's my own little interpretation about why the Dirac field should be "quantified" (i.e be an operator instead of an ordinary wave function), without all the talk about Pauli principle, anihilation/creation operators and all other weird QFT stuff. For me, it's enough to understand why $\Psi$ below (not $\psi$) isn't a probability amplitude.
In a relativistic setup, all fundamental particles and fields in nature should be irreducible representations of the Lorentz group. So we could have scalar, vector, tensor and also spinor fields.
Since any field should propagate in empty space without violating causality, the field should obey a wavelike differential equation (Klein-Gordon equation if the field has inertia ; $m \ne 0$). In the case of the electromagnetic field $F^{ab}$ (or the vector field $A^a$) propagating in empty space, it obeys the Maxwell equations, which includes the wave equation (with $m = 0$) :
$$\tag{1}
\partial_a \, F^{ab} = 0, \qquad \Rightarrow \qquad \square \, A^a = 0.
$$
In the case of a spinor field $\Psi$, it should be the Dirac first order equation, which is also including the wave equation (with a mass term) :
$$\tag{2}
\gamma^a \, \partial_a \, \Psi + i \, m \, \Psi = 0, \qquad \Rightarrow \qquad \square \, \Psi + m^2 \, \Psi = 0.$$
Now, all the fields are in principle obervables quantities, or could have observable effects (else, it's not physics !). The electromagnetic field is not directly obervable as such, but could have effects on electrical charges (which could reveal the presence of the electromagnetic field). In principle, the energy-momentum tensor $T^{ab}$ of the EM field is also obervable/measurable (since it's about energy, momentum, angular momentum, etc). So the $A^a$ field should be treated as an observable in quantum mechanics, which imposes all observables to be representend by hermitian operators.
It's the same for the spinor field $\Psi$. It is not directly obervable as such, but it could react to an electromagnetic field, and could also generate some EM field (if $\Psi$ has a charge). Its energy momentum $T^{ab}$ is also observable, in principle (energy, momentum, angular momentum and so on). So in Quantum Mechanics, it should be represented by an operator.
In a general case, you have a physical field $\Phi$ (indices suppressed) propagating in spacetime as an irreducible representation of the Lorentz group, thus obeying some partial differential equation of the general shape
$$\tag{3}
\mathcal{E}(\Phi, \; \partial_a \, \Phi, \; \partial_a \, \partial_b \, \Phi) = 0.
$$
If it has an energy momentum tensor $T^{ab}$ (typically depending on the squares $\Phi^2$ and $(\partial_a \Phi)(\partial_b \, \Phi)$), then it should be regarded as an observable in QM. This implies that it should be defined as an operator object. Not a probability amplitude. This is a very general consequence of standard QM, and has nothing to do with "second quantification".
"Second quantification" is really an historical mistake, done at an epoch where there was a lot of confusion about the fields and particles of Nature. It just happened that we discovered the electron firstly as a particle (i.e. interacting with measuring devices in the lab as "particles"), and "rediscovered" a bit later that it was really just another field propagating in spacetime. Electron isn't a pointlike particle! If you think a bit about it, fundamental pointlike particles simply doesn't make any physical sense at all.
There is no real particles out there. Just mathematical fields (i.e. representations of the Lorentz Group constrained by the causality principle) propagating like waves and interacting with other fields like particles.
