This is at the current limit of my understanding of fermionic QFT. It's such a good question that it deserves an answer from an expert.
a) Fourier analysing the second quantized Dirac operator field gives, $\hat{\psi}^{a}(x)$,
\begin{equation}
\hat{\psi}^{a}(x)=\int \frac{d^{3}p}{(2\pi)^{3/2}}\exp(ix^{r}p^{r})\hat{\psi}^{a}(p)
\end{equation}
where the label $a=1,2,3,4$ runs over the four components of the Dirac spinor and the label $r=1,2,3$ runs over the three spatial coords and I'm working at a constant time. So, borrowing Lior's introductory remarks, "The quantized Dirac field $\hat{\psi}^{a}(x)$ at a certain spatial point can be written as a linear combination of creation operators $\hat{\psi}^{a}(p)$ acting on the Hilbert space of physical states, with coefficients that are free field harmonic solutions $\exp(ix^{r}p^{r})$." I don't think the free field harmonic solutions are Dirac spinors because the spinor index is on the creation/annihilation operators. A state $|\Psi\rangle$ will be a sum of polynomials in the creation operators applied to a vacuum state $|S\rangle$,
\begin{equation}
|\Psi\rangle= (\ldots +c\hat{\psi}^{a}(p_{1})\hat{\psi}^{b}(p_{2})...\hat{\psi}^{c}(p_{n})+\ldots )|S\rangle
\end{equation}
The field operator $\hat{\psi}^{a}(x)$ applied to the state changes the state by modifying the polynomial just by multiplication.
b) The mathematical framework is the space of polynomials in the variables $\hat{\psi}^{a}(p)$.
c) The vacuum state $|S\rangle$ is the trivial polynomial $1$.
d) The Hamiltonian for the free electron field is,
\begin{equation}
\hat{H}=\int d^{3}x (-i\hat{\psi}^{\dagger}\gamma^{0}\gamma^{r}\frac{\partial\hat{\psi}}{\partial x^{r}}+m\hat{\psi}^{\dagger}\gamma^{0}\psi)
\end{equation}
where I've suppressed the spinor indices and so I'm using matrix notation for anything to do with the spinors. The chirality operator is $\gamma^{5}$.
\begin{equation}
\gamma^{5}=
\left[
\begin{array}{cc}
1 & 0 \\
0 & -1
\end{array}
\right]
\end{equation}
where the entries are 2x2 block matrices. $\gamma^{0}$ is,
\begin{equation}
\gamma^{0}=
\left[
\begin{array}{cc}
0 & 1 \\
1 & 0
\end{array}
\right]
\end{equation}
The action of the chirality operator is to change the Dirac spinors $\hat{\psi}\rightarrow \gamma^{5}\hat{\psi}$. Nothing happens to the gamma matrices. For simplicity, just look at what happens to the mass term.
\begin{equation}
\hat{\psi}^{\dagger}\gamma^{0}\hat{\psi}\rightarrow (\gamma^{5}\hat{\psi})^{\dagger}\gamma^{0}\gamma^{5}\hat{\psi}=\hat{\psi}^{\dagger}(\gamma^{5}\gamma^{0}\gamma^{5})\hat{\psi}=\hat{\psi}^{\dagger}(-\gamma^{0})\hat{\psi}=-\hat{\psi}^{\dagger}\gamma^{0}\hat{\psi}
\end{equation}
So, the mass term changes sign under the chirality transformation so the Hamiltonian is not invariant. Although the Hamiltonian looks like a "scalar" because we've summed over all the spinor indices, transforming each component of the spinor gives a result which is minus the thing we started with; so the Hamiltonian changes under chirality.
