Well, for $s\in G$ let $\lambda(s)$ be the left-translation operator by $s$; all such operators are in the group von Neumann algebra. I guess that the hoped for homomorphism $F:NG \rightarrow NH$ should satisfy $F(\lambda(s)) = \lambda(f(s))$ for $s\in G$, and we should have that $F$ is an (ultraweakly) continuous $*$-homomorphism. In particular, $F$ is contractive.

Then $F$ need not exist. Let $G=\mathbb Z$ and $H=\mathbb Z/n\mathbb Z$. Then $NH = CH$, so we have a trace on $NH$ which sends $\lambda(0)$ to $1$. So if we apply $F$, and then take the trace, we should get an ultraweakly continuous functional $\phi$ on $NG$ which satisfies $\phi(\lambda(ns)) = 1$ for all $s\in\mathbb Z$.

But this can't happen: maybe we can see this via the Fourier transform. Then $NG \cong L^\infty(\mathbb T)$ and $\phi$ induces $h\in L^1(\mathbb T)$ which satisfies $\int h(\theta) e^{ins\theta} d\theta = 1$ for all $s\in\mathbb Z$. This violated Reimann-Lebesgue.

On the other hand, if $G \subseteq H$ is an inclusion (of discrete groups, to avoid topology) then we do get an inclusion $NG \rightarrow NH$. Here's a construction. Find an index set $I$ and $(h_i)$ in $H$ such that $H$ is the disjoint union of $\{Gh_i\}$. Then define $V:l^2(H)\rightarrow l^2(G)\otimes l^2(I)$ by $V(\delta_h) = \delta_g\otimes\delta_i$ if $h=gh_i$ (so defined on point-masses, and extend by linearity). So $V$ is unitary, and $\theta:x\mapsto V^*(x\otimes 1)V$ is a normal $*$-homomorphism $NG\rightarrow B(l^2(H))$. Then, for $r\in G$, $V^*(\lambda(r)\otimes 1)V(\delta_h) = V^*(\delta_{rg}\otimes\delta_i) = \delta_{rh}$ as $rg\in G$. So $\theta$ maps into $NH$, and does what we want.

Surely there is some general result, but I'm not sure of it...