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).
Let me assume only that $G$ is a connected Lie group because simple connectedness does not seem to play any role. 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 $L_a$ be the Zariski closure of the image of the representation $j:G\to\mathrm{GL}(V)$; hence $L_a$ is Zariski connected. Write $H_a=(L_a)_v$ (the stabilizer of $v$), so $H_a$ is Zariski-closed in $L_a$. Then $G_v=j^{-1}(H_a)$.
So now we have: a connected Lie group $G$, a continuous homomorphism $j$ with discrete kernel and Zariski-dense image into a real connected linear algebraic group $L_a$, a Zariski-closed subgroup $H_a$ in $L_a$, and we wish to understand the group of components of $j^{-1}(H_a)$. The group of real points of $L_a$ has finitely many components in the Lie topology; let $L$ be the unit component, and let $H=L\cap H_a$. Since $j(G)\subset L$, we have $j^{-1}(H)=j^{-1}(H_a)$.
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 $[L_a,L_a]$.)
Let $p$ be the projection $L\to L/L'$. Then $L/L'$ is an abelian 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 $M$, which is normal in $j^{-1}(H)$. Then $j(M)$ is the unit component of $L'\cap H$ (in the Lie topology), so $\mathrm{Ker}(j)M$ has finite index in $j^{-1}(L'\cap H)$.
To summarize, we have the inclusions $\mathrm{Ker}(j)M\subset j^{-1}(L'\cap H)\subset j(H)$; these are closed normal subgroups, $M$ is connected, $\mathrm{Ker}(j)$ is central, the quotient $j^{-1}(L'\cap H)/\mathrm{Ker}(j)M$ is finite and the quotient $j^{-1}(H)/j^{-1}(L'\cap H)$ is abelian. Let $p$ be the projection $j^{-1}(H)\to \Gamma=\pi_0(j^{-1}(H))$. Then $p(M)$ is trivial, so $p(\mathrm{Ker}(j)M)=p(\mathrm{Ker}(j))$ is central; the quotient $p(j^{-1}(L'\cap H))/p(\mathrm{Ker}(j)M)$ is finite and the quotient $p(j^{-1}(H))/p(j^{-1}(L'\cap H))$ is abelian. Thus $\Gamma$ is central-by-(finite-by-abelian). [Since central-by-finite implies finite-by-abelian, $\Gamma$ is also central-by-(2-step-nilpotent)].
Proof of the claim that the Lie algebra $\mathfrak{p}$ of $j(G)$ is a ideal in the Lie algebra $\mathfrak{l}$ of $L$ and $\mathfrak{l}/\mathfrak{p}$ is abelian:
First, $j(G)$ normalizes $\mathfrak{p}$; since the normalizer in $L$ of $\mathfrak{p}$ is Zariski-closed and contains the Zariski-dense subgroup $j(G)$, it is equal to $L$. So $\mathfrak{p}$ is an ideal. Next, we use a theorem due to Chevalley (see e.g., Chevalley, Théorie des groupes de Lie, Hermann; Chap. II.14), which says that for any Lie subalgebra of $\mathfrak{l}$, its derived subalgebra is the Lie algebra of a Zariski-closed subgroup; thus let $Q$ be a Zariski-closed subgroup whose Lie algebra is $[\mathfrak{p},\mathfrak{p}]$. Then it follows that the derived subgroup of $[j(G),j(G)]$ is equal to the unit component of $Q$ (in the Lie topology). In particular, $[g,h]\in Q$ for any $g,h\in j(G)$. By Zariski density, it follows that $[L,L]\subset Q$. Thus $[\mathfrak{l},\mathfrak{l}]=[\mathfrak{p},\mathfrak{p}]\subset\mathfrak{p}$.
Update. I here answer the question in the comment: yes, every f.g. finite-by-abelian group occurs. Every such group $P$ can be written as fibered product $P=F\times_\Lambda\Gamma$, where $\Gamma$ is a free abelian group of finite rank, $F$ is a finite group, and $\Lambda$ is a common quotient (thus a finite abelian group). Choose $n$ and an embedding $F\to S=\mathrm{SL}_n(\mathbf{C})$, and an embedding of $\Lambda$ into a complex torus $T$ (defining by composition a map $F\to T$). Now embed $F$ diagonally into $S\times T$. Then the fundamental group of $(S\times T)/F$ is isomorphic to $P$ (see Corollary 3.1 of this note of mine with Borovoi; possibly there are earlier references). Finally use the theorem you refer to so as to realize $F$ as stabilizer of some vector for some $(S\times T)$-module.
Best Answer
Yes. Here is one way: Consider standard $\mathbb{R}^3$ endowed with the Lorentzian quadratic form $Q = x^2+y^2-z^2$, and let $G\simeq\mathrm{O}(2,1)\subset\mathrm{GL}(3,\mathbb{R})$ be the symmetry group of $Q$. Then $G$ preserves the hyperboloid $H$ of $1$-sheet given by the level set $Q=1$, which is diffeomorphic to a cylinder. Consider the quotient of $H$ by $\mathbb{Z}_2$ defined by identifying $v\in H\subset\mathbb{R}^3$ with $-v$. This abstract quotient is a smooth Möbius strip.
This quotient can be identified as a linear group orbit as follows: Let $V = S^2(\mathbb{R}^3)\simeq \mathbb{R}^6$ and consider the smooth mapping $\sigma:\mathbb{R}^3\to V$ given by $\sigma(v) = v^2$ for $v\in\mathbb{R}^3$. Then $\sigma$ is a $2$-to-$1$ immersion except at the origin. The action of $G$ on $\mathbb{R}^3$ extends equivariantly to a representation $\rho:G\to \mathrm{Aut}(V)$ such that $\rho(g)(v^2) = \rho\bigl(\sigma(v)\bigr)=\sigma(g v)= (gv)^2$. It follows that $\sigma(H)\subset S^2(\mathbb{R}^3)\simeq\mathbb{R}^6$, which is a Möbius strip, is a linear group orbit under the representation $\rho$.
Note that the representation of $G$ on $S^2(\mathbb{R}^3)\simeq\mathbb{R}^6$ is actually reducible as the direct sum of a trivial $\mathbb{R}$ and an irreducible $\mathbb{R}^5$. Projecting everything into the $\mathbb{R}^5$ factor, one obtains a representation of $G$ on $\mathbb{R}^5$ that has a Möbius strip as a $G$-orbit.