Indeed, the ordinary Schrödinger equation can be second-quantized, yielding non-relativistic QFT.
To do so, one first rewrites the Schrödinger equation as a "classical" field theory:
the phase space of this field theory is the Hilbert space $\mathcal{K}$ of the first-quantized theory
with (complex) coordinates $a_k(\psi) = \left\langle e_k \middle| \psi \right\rangle$ for some ONB $\left(e_k\right)_k$ of $\mathcal{K}$
obeying the Poisson brackets $\left\{ a_k, a^*_l \right\} = -i \delta_kl$
and with Hamiltonian function(al) $\mathcal{H}(\psi) = \left\langle \psi \middle| H \middle| \psi \right\rangle$, where $H$ denotes the first-quantized Hamiltonian operator (that this Hamiltonian indeed encodes the first-quantized Schrödinger equation as can be checked from $\frac{d}{dt} a_k(\psi) = \{ a_k, \mathcal{H} \}$).
Remarkably, the canonical (second-)quantization of this field theory describes precisely an arbitrary number of undistinguishable (bosonic) particles, which each obey the first-quantized theory:
the second-quantized Hilbert space is the Fock space $\mathcal{F} = \bigoplus_{N \geq 0} \mathcal{K}^{\otimes N, \text{sym}}$, where $\mathcal{K}^{\otimes N, \text{sym}}$ is the symmetric subspace of $\mathcal{K}^{(1)} \otimes \dots \otimes \mathcal{K}^{(N)}$, which describes $N$ undistinguishable particles
the quantization of $a_k$, resp. $a^*_l$, is the annihilation operator $\hat{a}_k$ for a particle of "type $e_k$", resp. the creation operation $\hat{a}^+_l$, whose commutator is indeed $\left[ \hat{a}_k, \hat{a}^+_l \right] = \delta_kl$
the second-quantized Hamiltonian, obtained by quantizing $\mathcal{H}$, is $\hat{H} = \bigoplus_{N \geq 0} \sum_{n=1}^N 1^{(1)} \otimes \dots \otimes H^{(n)} \otimes \dots \otimes 1^{(N)}$ (this is easier to show if the ONB $\left(e_k\right)_k$ is chosen as the eigenbasis of the first-quantized Hamiltonian $H$, in which case $\hat{H} = \sum_k E_k \hat{a}^+_k \hat{a}_k$), meaning that each particle evolves independently, according to the first-quantized Hamiltonian H.
Of course, the Schrödinger equation is non-relativistic, so if $e_k$ is an energy eigenstate for, say, a single electron, the corresponding $E_k$ will correspond to the NR limit $E = \sqrt{m^2 c^4 + p^2 c^2} \approx m c^2 + \frac{p^2}{2m}$. Although the constant term $mc^2$ is often dropped from the 1st quantized theory, one should keep it here if we want the creation/annihilation operator to change the total energy by the correct order of magnitude.
I like this example as it cleanly demonstrates how the 2 aspects of second-quantization tie together, namely:
describing an arbitrary number of particles obeying this first-quantized theory;
taking a (first-quantized) wave-equation, and thinking of it as a "classical" field theory to be quantized again, hence the name "second-quantization": although the physical interpretation is a priori different, the mathematics of going from the 1st to the 2nd-quantized theory happen to be exactly the same as the ones to go from the classical theory to the 1st-quantized one!
In a typical introduction to QFT, which goes directly to the relativistic case and deals with the second-quantization of the Dirac/Klein-Gordon wave-equations, this remarkable match is somewhat obscured by the lack of a well-defined first-quantized theory (due to the pathologies of these relativistic wave-equations).
In addition, the above holds for any first-quantized theory, so one can see the magic happens already by working out the second-quantization of a simple spin $1/2$: there, the phase space to be (second-)quantized is just 4 dimensional (thus bypassing field theoretic subtleties arising from an infinite-dimensional phase space), with two independent complex variables, each of which can be identified with the $z = x + ip$ variable of an harmonic oscillator. For reference purposes, the more difficult case of 2nd-quantizing the Schrödinger equation of a non-relativistic particle is treated in details eg. in Schiff L.I. "Quantum mechanics" book (sections 45 & 46 of chapter XIII).
@Andrew's answer provides the big picture, but I'd like to give a few more specific pointers that hopefully may help.
Questions 1-2: Does everything look correct so far? How to express these operators in terms of one-particle operators?
So you want to set up single-particle analogues of the ladder operators using eigenstates of the 1st quantized Hamiltonian, then use those to construct the second quantization ladder operators in the symmetrized multi-particle space. The problem is that such a procedure may not be possible. Here is why:
The entire second quantization framework rests on an isomorphism between the (anti)symmetric subspace of the N-particle Hilbert space and an abstract direct product of "mode" Hilbert spaces, each constructed around its own ladder operator algebra. The ladder operators ${\hat a}_n$, ${\hat a}^\dagger_n$ in the abstract/"mode" Hilbert space obviously do have equivalents on the original N-particle (anti)symmetric subspace. But cannot be expressed as symmetrized sums of similar single-particle operators. To see why, let us suppose the ${\hat a}_n$, ${\hat a}^\dagger_n$ can indeed be expressed as such symmetrized sums, reading
$$
{\hat a}_n \sim \sum_{k=1}^N{ {\hat \alpha}^{(k)}_n},\;\;\; {\hat a}^\dagger_n \sim \sum_{k=1}^N{ \left( {\hat \alpha}^{(k)}_n\right)^\dagger}
$$
up to some suitable normalization factor, where ${\hat \alpha}^{(k)}_n$, $\left( {\hat \alpha}^{(k)}_n\right)^\dagger$ are the desired "single-particle ladder operators" for particle $k$ and "mode"/eigenstate $n$, and each term is to be understood in the sense of ${\hat \alpha}^{(k)}_n\otimes \left[\bigotimes_{j\neq k}{\hat I}^{(j)}\right]$. The latter can be naturally assumed to (anti)commute for different particles and eigenstates (pre-symmetrization), so $\left[{\hat \alpha}^{(k)}_m, {\hat \alpha}^{(j)}_n\right]_\pm = \left[{\hat \alpha}^{(k)}_m, \left( {\hat \alpha}^{(j)}_n\right)^\dagger\right]_\pm = 0$ for any $j\neq k$ and for $n\neq m$ when $j = k$. Then the (anti)commutation relations for the ${\hat a}_n$-s,
$$
\left[{\hat a}_m, {\hat a}_n\right]_\pm = 0,\;\;\; \left[ {\hat a}_m, {\hat a}^\dagger_n\right]_\pm = \delta_{mn}{\hat I}
$$
require
$$
0 = \left[{\hat a}_m, {\hat a}_n\right]_\pm \sim \sum_{k=0}^N{\left[{\hat \alpha}^{(k)}_m, {\hat \alpha}^{(k)}_n\right]_\pm }, \;\;\;\;
\delta_{mn}{\hat I} = \left[{\hat a}_m, {\hat a}^\dagger_n\right]_\pm \sim \sum_{k=0}^N{\left[{\hat \alpha}^{(k)}_m,\left( {\hat \alpha}^{(k)}_n\right)^\dagger\right]_\pm }
$$
The important point here is that in either case the lhs does not depend on $N$. Then the first eq. above implies that each term on the rhs must vanish identically, which is great. But things are no longer clear cut for the 2nd eq. Whatever prescription we might propose for $\left[{\hat \alpha}^{(k)}_m,\left( {\hat \alpha}^{(k)}_n\right)^\dagger\right]_\pm $, there is no way to normalize the sum such that the result is non-zero yet independent of $N$ for any $N$.
Bottom line: however intuitive it may seem at first sight, this is not the way to go.
Question 3: In QFT why do we treat those excitations as multi-particle states?
Short answer: Due to the isomorphism with the many-particle framework. Sometimes, as in solid state physics, the excitations are referred to as quasi-particles for this exact reason. Think phonons and excitons. Same goes for photons and any other field quanta, but for historical reasons they are referred to as "particles".
Question 4: Can we 'second-quantise' the EM field by treating the Maxwell equations as a Schrödinger equation (e.g. see books by Fushchich and Nikitin) and then considering the multi-particle states of those?
Unfortunately I'm not familiar with the book you mention, but you may want to google "Maxwell equations in Dirac form", for instance this paper and this Wikipedia page (especially refs. within). Never saw this used as a starting point for second quantization though, but why not? Perhaps an interesting idea?
Question 5: Why are the two procedures equivalent and lead to the same result?
Another short answer, same as for Question 3: Because both procedures rely on an isomorphism with the same type of abstract Hilbert space and the associated operator algebra. As mentioned before, the "Classical Mechanics"/"particle" procedure observes an isomorphism between the finite (anti)symmetric subspace of the N-particle Hilbert space and a "fixed particle number" subspace of the abstract second quantized space, while the "Field Theory" procedure yields a true isomorphism (the "postulate" you mention).
Best Answer
$\renewcommand{ket}[1]{|#1\rangle}$ Item #4 in your list is best thought of as the definition of the word "particle".
Consider a classical vibrating string. Suppose it has a set of normal modes denoted $\{A, B, C, \ldots\}$. To specify the state of the string, you write it as a Fourier series
$$f(x) = \sum_{\text{mode } n=\in \{A,B,C,\ldots \}} c_n [\text{shape of mode }n](x) \, .$$
In the typical case, $[\text{shape of mode }n](x)$ is something like $\sin(n\pi x / L)$ where $L$ is the length of the string. Anyway, the point is that you describe the string by enumerating its possible modes and specifying the amount by which each mode is excited by giving the $c_n$ values.
Suppose mode $A$ has one unit of energy, mode $C$ has two units of energy, and all the other modes have zero units of energy. There are two ways you could describe this situation.
Enumerate the modes (good)
The first option is like the Fourier series: you enumerate the modes and give each one's excitation level: $$|1\rangle_A, |2\rangle_C \, .$$ This is like second quantization; we describe the system by saying how many units of excitation are in each mode. In quantum mechanics, we use the word "particle" instead of the phrase "unit of excitation". This is mostly because historically we first understood "units of excitations" as things we could detect with a cloud chamber or Geiger counter. To be honest, I think "particle" is a pretty awful word given how we now understand things.
Label the units of excitation (bad)
The second way is to give each unit of excitation a label, and then say which mode each excitation is in. Let's call the excitations $x$, $y$, and $z$. Then in this notation the state of the system would be $$\ket{A}_x, \ket{C}_y, \ket{C}_z \, .$$ This is like first quantization. We've now labelled the "particles" and described the system by saying which state each particle is in. This is a terrible notation though, because the state we wrote is equivalent to this one $$\ket{A}_y, \ket{C}_x, \ket{C}_z \, .$$ In fact, any permutation of $x,y,z$ gives the same state of the string. This is why first quantization is terrible: particles are units of excitation so it is completely meaningless to give them labels.
Traditionally, this terribleness of notation was fixed by symmetrizing or anti-symmetrizing the first-quantized wave functions. This has the effect of removing the information we injected by labeling the particles, but you're way better off just not labeling them at all and using second quantization.
Meaning of 2$^{\text{nd}}$ quantization
Going back to the second quantization notation, our string was written $$\ket{1}_A, \ket{2}_C$$ meaning one excitation (particle) in $A$ and two excitations (particles) in $C$. Another way to write this could be to write a single ket and just list all the excitation numbers for each mode: $$\ket{\underbrace{1}_A \underbrace{0}_B \underbrace{2}_C \ldots}$$ which is how second quantization is actually written (without the underbraces). Then you can realize that $$\ket{000\ldots \underbrace{N}_{\text{mode }n} \ldots000} = \frac{(a_n^\dagger)^N}{\sqrt{N!}} \ket{0}$$ and just write all states as strings of creation operators acting on the vacuum state.
Anyway, the interpretation of second quantization is just that it's telling you how many excitation units ("quanta" or "particles") are in each mode in exactly the same way you would do it in classical physics.
See this post.
Comments on #4 from OP
In introductory quantum we learn about systems with a single particle, say, in a 1D box. That particle can be excited to a variety of different energy levels denoted $\ket{0}, \ket{1},\ldots$. We refer to this system as having "a single particle" regardless of which state the system is in. This may seem to run contrary to the statements made above in this answer in which we said that the various levels of excitation are referred to as zero, one, two particles. However, it's actually perfectly consistent as we now discuss.
Let's write the equivalent first and second quantized notations for the the single particle being in each state: $$\begin{array}{lllll} \text{second quantization:} & \ket{1,0,0,\ldots}, & \ket{0,1,0,\ldots}, & \ket{0,0,1,\ldots} & \ldots \\ \text{first quantization:} &\ket{0}, &\ket{1}, &\ket{2}, & \ldots \end{array} $$ Although it's not at all obvious in the first quantized notation, the second quantized notation makes clear that the various first quantized states involve the particle occupying different modes of the system. This is actually pretty obvious if we think about the wave functions associated to the various states, e.g. using first quantized notation for a box of length $L$ \begin{align} \langle x | 0 \rangle & \propto \sin(\pi x / L) \\ \langle x | 1 \rangle & \propto \sin(2\pi x / L) \, . \end{align} These are just like the various modes of the vibrating string. Anyway, calling the first quantized states $\ket{0}$, $\ket{1}$ etc. "single particle states" is consistent with the idea that a particle is a unit of excitation of a mode because each of these states has one total excitation when you sum over all the modes. This is really obvious in second quantized notation.