Meaning of Fock Space
Suppose you have a system described by a Hilbert space $H$, for example a single particle. The Hilbert space of two non-interacting particles of the same type as that described by $H$ is simply the tensor product
$$H^2 := H \otimes H$$
More generally, for a system of $N$ particles as above, the Hilbert space is
$$H^N := \underbrace{H\otimes\cdots\otimes H}_{N\text{ times}},$$
with $H^0$ defined as $\mathbb C$ (i.e. the field underlying $H$).
In QFT there are operators that intertwine the different $H^N$s, that is , create and annihilate particles. Typical examples are the creation and annihilation operators $a^*$ and $a$. Instead of defining them in terms of their action on each pair of $H^N$ and $H^M$, one is allowed to give a "comprehensive" definition on the larger Hilbert space defined by taking the direct sum of all the multi-particle spaces, viz.
$$\Gamma(H):=\mathbb C\oplus H\oplus H^2\oplus\cdots\oplus H^N\oplus\cdots,$$
known as the Fock Hilbert space of $H$ and sometimes also denoted as $e^H$.
From a physical point of view, the general definition above of Fock space is immaterial. Identical particles are known to observe a definite (para)statistics that will reduce the actual Hilbert space (by symmetrisation/antisymmetrisation for the bosonic/fermionic case etc...).
Great answers, but just for completeness maybe it will be illustrative to have an example.
Suppose your $H^1$ contains some single-particle states $|a\rangle$, $|b\rangle$, etc. The Fock space removes the limitation on being a single particle, and is composed of $H^0$ (which is 1-dimensional), $H^1$, $H^2 = H \otimes H$, etc. This allows states like
- the vacuum state, let's call it the empty ket $|\rangle$,
- all single particle states, $|a\rangle, |b\rangle, \ldots$,
- all two-particle states, $|aa\rangle, |ab\rangle, |ba\rangle, \ldots$ (NB that this construction deems them distinguishable),
but most importantly
- any superposition of the above, like $\frac{e^{i\pi/4}}{\sqrt2}|\rangle + \frac12 |a\rangle - \frac12|aab\rangle\otimes\left(\frac1{\sqrt2}|a\rangle + \frac i{\sqrt2}|b\rangle\right)$.
This space is inherently infinite-dimensional even if you start with something small like a qubit. If you want to imagine the result with the help of a basis, simply concatenate the lists of the basis states of all the components:
$$\{|\rangle, |0\rangle, |1\rangle, |00\rangle, |01\rangle, |10\rangle, |11\rangle, |000\rangle, |001\rangle, \ldots\}$$
In the most trivial setting the single particle does not really have any distinct states, so $H^1$ is 1-dimensional. It still makes sense to pick a fiducial state $|{}\circ{}\rangle \in H^1$ and construct the Fock space with basis
$$\{|\rangle =: |0\rangle, |{}\circ{}\rangle =: |1\rangle, |{}\circ{}\circ{}\rangle =: |2\rangle, |{}\circ{}\circ{}\circ{}\rangle =: |3\rangle, \ldots\},$$
an example of a state might be, say, a coherent state
$$|\alpha\rangle = \sum_{n=0}^{\infty} \frac{\alpha^n}{\sqrt{e^{|\alpha|^2}n!}} |n\rangle$$
and you have a nice example of why people can speak of excitations as of "phonons" in a harmonic oscillator even though there's just a single particle oscillating!
Yes, it does. You build a "large" Hilbert space from the "small" ones, if you like.