Since $G$ is a Lie group with finitely many components, it has a maximal compact subgroup $K$, unique up to conjugation and $G=KG_0$ (Mostow). Since $G$ is actually nilpotent, $K$ is unique and actually consists of the elements in $G$ contained in a compact subgroup. In particular $K$ is normal in $G$.

From $G$ we can define $K$ as above; conversely if we know $K$ and $G_0$, we know $G$, which is the quotient of $K\times G_0$ by the central subgroup $\{(g,g^{-1}),g\in K_0\}$.

This shows that if $F$ is a finite nilpotent group, the set classifying extensions of $G_0$ by $F$ that are nilpotent is in canonical bijection with the set classifying central extensions of $K_0$ by $F$. The latter is in bijection with $H^2(F,K_0)$, which is a reasonable well-understood object ($K_0$ is a torus).

In particular:
1) If $G_0$ is simply connected we get $G=G_0\times F$ (only the direct extension)

2) Everything boils down to when $G_0$ is a torus. If $S^1$ is the 1-circle, $Q$ is the group of quaternions of order 8 and if we define $G$ as the quotient of $S^1\times Q$ by the diagonal element of order 2, then we get a nontrivial extension (as we see by counting elements of order 2).

NB: you assume explicitly that $G$ is nilpotent. So I don't claim to classify all extensions of $G_0$ by $F$.