It is conjectured that no such example exists. It is conjectured that $G$ is of type $F$, but it is hard to prove a group is type $F$ without explicitly exhibiting a classifying space. The class of groups of type $FP$ is a well-behaved proxy. Moreover, it is conjectured that every finitely presented group of type $FP$ is actually of type $F$. It is known that a finitely presented group of type $FP$ has a classifying space which is "finitely dominated" meaning that it is a retract in the homotopy category of a finite complex, which is just as good for most homotopical purposes. I will prove that if a torsion-free group $G$ has a finite index subgroup $H$ of type $FP$, then $G$ is itself of type $FP$. Applying this to your situation, since $H$ is finitely presented, so is $G$, so the conjecture implies that $G$ is of type $F$. It is possible that the hypothesis that $H$ is of type $F$ allow an unconditional proof, but I do not expect it.

$G$ has type $FP$ if the trivial $G$-module $\mathbb Z$ has a finite length resolution by finitely generated projective $G$-modules. Type $F$ is equivalent to a finite length resolution by finitely generated free $G$-modules, plus finite presentation. The advantage of type $FP$ is that it is the intersection of two properties that can be treated separately, namely finite dimension and finite generation of cohomology $FP_\infty$. These can be interpreted as the existence of two resolutions, one of finite length and other by finite modules. The two properties can be combined to produce a resolution that is simultaneously finitely generated and finite length, but only if one allows projective modules and not just free modules, so type $F$ does not decompose this way.

The easy part is finite generation of homology. If $G$ contains a finite index subgroup $H$, then the intersection of the conjugates of $H$ is a finite index normal subgroup $N$. The Hochschild-Serre spectral sequence for the group extension $N\to G\to G/N$ allows us to combine the $FP_\infty$ properties of $N$ and $G/N$ to conclude that $G$ is $FP_\infty$.

The other part is finite homological dimension. By assumption $H$ has a finite dimensional classifying space with universal cover $EH$. Consider the space of $H$-equivariant maps from $G$ into $EH$. That is, intertwining the left action of $H$ on $G$ and the only action of $H$ on $EH$. This leaves the right action of $G$ on $G$, which makes this a $G$-space, a candidate for $EG$. For the quotient to be a model of $BG$, we need the space to be contractible and the action to be free. This space is isomorphic to $EH^{[G:H]}$, hence contractible and finite dimensional. The size of the isotropy is bounded by the index. So the isotropy is a finite subgroup of the torsion-free group $G$, so actually $G$ acts freely on this space. This shows that $G$ has a finite dimensional classifying space. In fact, the dimension of $G$ is the same as the dimension of $H$, although that does not come out of this construction.

Reference: Ken Brown's book, Chapter VIII or his notes.