I think we only obtain central-by-finite-by-abelian finitely generated groups (I don't claim we get all of them; maybe we only get finite-by-abelian groups and this is what I expect).

(I use the convention that a group is P-by-Q if it lies in an exact sequence with kernel satisfying P and quotient satisfying Q.)

This is enough to discard $\mathrm{SL}_2(\mathbf{Z})$, torsion-free nilpotent groups of nilpotency class $\ge 3$, or even many virtually abelian groups such as the infinite dihedral group, but not the discrete Heisenberg group.

Actually, a central-by-finite-by-abelian f.g. group is either finite-by-abelian or contains a copy of the discrete Heisenberg group so in a sense it's the last guy to rule out (if what follows is correct).

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First let me assume only that $G$ is a connected Lie group because it's more flexible. Then clearly the question does not change if we mod out by the unit component of the kernel of the representation, so we can suppose that the representation has a discrete (hence central) kernel.

Let $L$ be the Zariski closure of the image of the representation $j:G\to\mathrm{GL}(V)$; hence $L$ is Zariski connected. Write $H=L_v$, so $H$ is Zariski-closed in $L$. Then $G_v=j^{-1}(H)$.

So now we have: a connected Lie group $G$, a continuous homomorphism $j$ with discrete kernel and Zariski-dense image into a real connected algebraic group $L$ (which I identity with its set of real points, which is a virtually connected Lie group), a Zariski-closed subgroup $H$ in $L$, and we wish to understand the group of components of $j^{-1}(H)$.

Since $j(G)$ is Zariski dense in $L$, its Lie algebra is an ideal in the Lie algebra of $L$ with abelian quotient. In particular (by Zariski connectedness of $L$) its Lie algebra is normalized by $L$, and thus by connectedness of $G$, we obtain that $j(G)$ is normal in $L$ with abelian quotient. In particular, the derived subgroup $L'$ of $L$ (viewed as the group of real points) is contained in $j(G)$. (This derived subgroup has finite index in the group of real points of the algebraic group $L$.)

Let $p$ be the projection $L\to L/L'$. Then $L/L'$ is an abelian virtually connected Lie group; $p(H)$ is closed in $L/L'$; the image $p\circ j(G)\subset L/L'$ is a connected immersed Lie subgroup, and hence $(p\circ j)^{-1}(p(H))/\mathrm{Ker}(p\circ j)$, as a closed subgroup of a connected abelian Lie group, is isomorphic to the direct product of a connected abelian Lie group and a free abelian group of finite rank.

By definition, we have $\mathrm{Ker}(p\circ j)\cap j^{-1}(H)=j^{-1}(L'\cap H)$, which equals $j^{-1}(L')\cap j^{-1}(H)$. Recall that $L'\cap H\subset L'\subset j(G)$, so its Lie algebra is the image by $j$ of some Lie subalgebra of the Lie algebra of $G$, generating an immersed subgroup $M$. Then $j(M)$ is the unit component of $L'\cap H$ (in the Lie topology), so $j^{-1}(L'\cap H)$ has finite index in $\mathrm{Ker}(j)M$.  

Thus $j^{-1}(H)$ is an iterated extension of $\mathrm{Ker}(j)M$, a finite group, and a compactly generated  abelian Lie group. Thus the group of components $\pi_0(j^{-1}(H))$ is extension of the central subgroup $\mathrm{Ker}(j)$ by a finite-by-abelian finitely generated group. [Since central-by-finite implies finite-by-abelian, this is also central-by-(2-step-nilpotent)].