First, to associate Hilbert spaces you should choose a proper time-like foliation of the space-time. Then each space-like hypersurface of the foliation will be described by quantum state in a given Hilbert space. In other words, Hilbert space are given for fixed time always. The time evolution will then act on observables or states of this Hilbert space.
Now, in local QFTs we have the following condition that can be imposed in local observables:
$$
\left[\mathcal{O}_i(x),\mathcal{O}_j(y)\right]=0
$$
for $x-y$ space-like. Remember that local observables are observables that behaves properly under Poincare transformation:
$$
U(\Lambda, b)\mathcal{O}_i(x)U(\Lambda, b)^{\dagger}=R_i\,^j\,\mathcal{O}_j(\Lambda x+b)
$$
where $R$ is a representation of the Lorentz symmetry. This is all you need to construct Hilbert spaces for each point $x$ of the space-like hypersurface: For each point in space $x$ we construct the Hilbert space $\mathcal{H}_x$ associated to a complete set of local observables at point $x$, namely $\{\mathcal{O}_{i}(x),...\}$.
To construct a Hilbert space of a given region $A$ of space, just make the tensor product of the Hilbert spaces $\mathcal{H}_x$'s for $x\in A$:
$$
\mathcal{H}_{A}=\bigotimes_{x\in A}\mathcal{H}_x
$$
What makes local QFT so special is that, in principle, all the observables in that theory could be constructed out of local fields $\phi(x)$, $\psi_{\alpha}(x)$, $A_{\mu}(x)$ and etc. So this means that the Hilbert space of the theory is given by
$$
\mathcal{H}=\bigotimes_{\forall x} \mathcal{H}_x
$$
This means that all the states of the theory is described by the superpositions you can make with theses copies $\mathcal{H}_x$. Eigenstates of non-local observables like the energy $H$ will be entangled states, e.g. the vacuum state $|0\rangle$. This is why accelerated observers sees the Unruh effect. Accelerated observer just have access to a region $A$ of the whole space $A\cup A^{c}$ due to the Rindler horizon, so you get a description:
$$
\rho_A= tr_{A^{c}} \{|0\rangle\langle 0|\}
$$
that is clearly a mixed state. Actually if you do all the computations right you get that for accelerated observer at acceleration $2\pi/\beta$ separate the vacuum state as:
$$
|0\rangle=\sum_{i}e^{-(1/2)\beta E_i}|E_i\rangle _A \otimes|E_i\rangle _{A^c}
$$
where $E_i$ is the eigenvalue for the energy $H_{A/A^{c}}$ restricted just to the region $A$ or $A^{c}$. Doing the trace of the region behind the Rindler horizon you will get that the density matrix is:
$$
\rho_A=\sum_{i}e^{-\beta E_i}|E_i\rangle \langle E_i|_A
$$
a density matrix of a canonical ensemble at temperature $\beta^{-1}$.
Note that all this is based on the fact that things like $\bigotimes_{x}\mathcal{H}_x$ makes sense. To make sense of it we need to regularize and renormalize it. As David M point out, this may be not possible. A obvious way to see if this is possible is search for a lattice regulator, then we simple make the tensor product of each cell.
