A partial answer:
Let $\psi \in \mathcal D(N)$; in particular, it follows that $\psi \in \mathcal D(N^{1/2})$. We want to show that these conditions are sufficient such that e.g. $a(f)\psi \in \mathcal D(N^{1/2})$. Then the product $a^\dagger(g) a(f) \psi$ would be well-defined.
To do so, I think we can use the fact that for all $\psi \in \mathcal D(N^{1/2})$ it holds that$^\ddagger$
$$||a(f)\psi||_F \leq ||f||_{\mathfrak h}\, ||N^{1/2}\psi||_F \quad ,\tag{1}$$
where $\mathfrak h$ denotes the underlying one-particle space and $f\in \mathfrak h$. We then proceed by noting that if $\psi=(\psi_n)_n \in \mathcal D(N)$ we have that
$$ \infty > ||f||_\mathfrak h^2 ||N\psi||_F^2 = ||f||_\mathfrak h^2\sum\limits_n n^2||\psi_n||^2 = ||f||_\mathfrak h ^2\sum\limits_n n \underbrace{||(N^{1/2}\psi)_n||^2}_{||N^{1/2}\psi_n||^2_F} \geq \sum\limits_n n ||(a(f)\psi)_n||_F^2 =||N^{1/2}a(f)\psi||_F^2\tag{2} $$
and hence $a(f)\psi\in \mathcal D(N^{1/2})$.
I further think that the domains $\mathcal D(N^{/2})$ for all $K\in \mathbb N$ are dense, since the dense subspace $F_{\mathrm{fin}}$ of $F$ which contains all vectors $\psi=(\psi_n)_n$ with only finitely many non-zero $\psi_n$ is a subspace of these domains.
It remains to show that $(1)$ holds in the said domain without using some result we want to prove. In Ref. 1. a proof is sketched, which seems plausible to me.