Broadly speaking I'd say your interpretation is basically right.
Consider a system of spinless, non-interacting fermions in a box with volume $V$. The vacuum state $|0\rangle$ is annihilated by all of the annihilation operators $a_{\mathbf k}$. In order to construct a Hilbert space for our theory, we first consider the vector space of objects $\mathscr V$ which can be obtained by applying a finite number of creation operators to $|0\rangle$, and linear combinations thereof. We can equip this vector space with an inner product which is defined via the canonical anticommutation relations:
$$\psi = a_{\mathbf k_1}^\dagger a_{\mathbf k_2}^\dagger |0\rangle \qquad \phi = a^\dagger_{\mathbf q_1} a^\dagger_{\mathbf q_2} |0\rangle \qquad \{a_\mathbf k,a_\mathbf q\}=\{a^\dagger_\mathbf k,a^\dagger_\mathbf q\}=0 \qquad \{a_\mathbf k,a^\dagger_{\mathbf q}\} = \delta_{\mathbf k\mathbf q}$$
$$\implies \langle \psi,\phi\rangle = \langle 0|a_{\mathbf k_2} a_{\mathbf k_1} a^\dagger_{\mathbf q_1} a^\dagger_{\mathbf q_2}|0\rangle = \delta_{\mathbf k_1 \mathbf q_1}\delta_{\mathbf k_2 \mathbf q_2}-\delta_{\mathbf k_1 \mathbf q_2}\delta_{\mathbf k_2 \mathbf q_1}$$
$V$ is not a Hilbert space yet because it is not toplogically complete, so our final step is to take the Hilbert space completion of it by "adding" the Cauchy sequences of elements of $V$. This space $\overline V$ is what Fradkin calls the physical Hilbert space built from $|0\rangle$.
If we equip this space with a (non-interacting) Hamiltonian $H = \sum_\mathbf k \epsilon(\mathbf k) a^\dagger_\mathbf k a_\mathbf k$, we may talk about the ground state of an $N$-particle system, which is the filled Fermi sea. Clearly this state can be obtained by acting on $|0\rangle$ with a finite number of creation operators; explicitly, it is given by
$$FS = \frac{1}{\sqrt{N}}\prod_{|\mathbf k|<k_F} a^\dagger_\mathbf k |0\rangle$$
If we'd like, we may start with $FS$ and try to build a Hilbert space over that instead. We define
$$b_\mathbf k := \begin{cases} a_\mathbf k & |\mathbf k|>k_F \\ a^\dagger_{-\mathbf k} & |\mathbf k|<k_F\end{cases}$$
which all annihilate $FS$. The original Hamiltonian can be re-expressed as
$$H = \sum_{\mathbf k} \epsilon(\mathbf k) a^\dagger_\mathbf k a_\mathbf k = \sum_{|\mathbf k|> k_F}\epsilon(\mathbf k)b^\dagger_\mathbf k b_\mathbf k - \sum_{|\mathbf k|<k_F}\epsilon(-\mathbf k)b^\dagger_\mathbf k b_\mathbf k + E_0$$
where $E_0 \equiv \sum_{|\mathbf k|<k_F} \epsilon(\mathbf k)$ is the ground state energy of the filled Fermi sea. The (positive-energy) excitations created above the Fermi surface are "genuine" particles, while the (negative-energy) excitations created below the Fermi surfaces are "holes."
It's fairly easy to see that the physical Hilbert space constructed from $|0\rangle$ is the same as the physical Hilbert space constructed from $FS$. However, we now consider the thermodynamic limit $N,V\rightarrow \infty$ such that $n\equiv N/V$ remains fixed. The problems with that arise immediately.
The Fermi momentum is implicitly defined such that $N = \sum_{|\mathbf k|<k_F}$. In the thermodynamic limit of large $V$ the sum can be converted to an integral, where we have
$$N = \sum_{|\mathbf k|<k_F}\approx \frac{V}{(2\pi)^3}\int_{|\mathbf k|<k_F} \mathrm d^3 k = \frac{V}{(2\pi)^3}\frac{4}{3}\pi k_F^2 $$
$$\implies k_F \propto \sqrt{N/V}$$
As a result, the Fermi momentum and energy are functions only of $n$, and don't change as we take the thermodynamic limit. On the other hand, $E_0 = \frac{V}{(2\pi)^3} \int_{|\mathbf k|<k_F} \epsilon(\mathbf k) \mathrm d^3k\rightarrow \infty$, making the re-expressed Hamiltonian ill-defined.
Additionally, $FS$ can no longer be obtained by acting on $|0\rangle$ with a finite number of creation operators, nor can it be written as the limit of a Cauchy sequence of such vectors. As a result, taking the thermodynamic limit decouples the physical Hilbert space over $|0\rangle$ from the physical Hilbert space over $FS$.
All in all, I have the feeling that the above quote concerns the possible unitary inequivalent representations of the canonical anti-commutation relations, but due to my limit background I cannot really make sense of this.
Yes, that's the idea. For finite $N$ and $V$, there are a countable number of allowed momentum modes, each with a countable occupation number (for fermions of course this occupation number is either 0 or 1, but the generalization is painless). Therefore, by enumerating these modes $\mathbf k_1,\mathbf k_2,\ldots$ we may specify states in using the occupation number representation $|n_1,n_2,\ldots\rangle$. The occupation number basis can be associated with the set $\mathbb N^\mathbb N$.
If we consider the linear span of the occupation number basis elements, we obtain a vector space $\mathscr S$ which is a strict superspace of the $\mathscr V$ described at the top of my answer. The reason for this is that $\mathscr V$ does not include basis vectors with an infinite number of nonzero occupation numbers, e.g. the vector $|1,1,1,\ldots\rangle$. In order to make a well-defined (i.e. finite) inner product, we first choose a unit norm "reference state" (such as $|0,0,0,\ldots\rangle$), then consider the vector space generated by vectors which differ from this in a finite number of places (our $\mathscr V$), equip it with an inner product derived from the CCR/CARs, and finally take the Hilbert space completion to define the physical Hilbert space.
When $N$ and $V$ are finite, $FS$ differs from $|0\rangle$ in a finite number of places (by definition) and therefore belongs to the same physical Hilbert space. In the thermodynamic limit, however, there are an uncountable number of momentum modes (so the occupation number basis set becomes isomorphic to $\mathbb R^\mathbb N$) and $FS$ does not differ from $|0\rangle$ in a finite number of places. In this case, the vector space $\mathscr S$ decomposes into an uncountable infinity of subspaces, each of which can be made into a physical Hilbert space in its own right.
