Let $L$ be a group. We say a group $H$ only admits finite images of $L$, if for every homomorphism $f\colon L\to H$ the image $f(L)$ is finite.
Now, assume that $G$ is a group and $H\subseteq G$ be a finite index subgroup of $G$, which only admits finite images of $L$.
Does $G$ only admit finite images of $L$?
Firstly, I noticed, that the "only admits finite images of $L$" condition is preserved under taking subgroups and finite direct products. So, I tried to show, that this condition is preserved under extensions, since every finite group admits this condition trivially and the group $G$ above is an extension of (a subgroup of) $H$ by some finite group.
For this, I tried using the universal embedding theorem by Krasner–Kaloujnine. So I can simplify the question to $G$ being a semidirect product of $H$ and a finite group, but now I am completely stuck.
I think this works to produce a counterexample. Hopefully no one will find an error...
Let $T_p$ be a Tarski monster of exponent $p$ with an outer automorphism of order $3$, with $3\nmid p-1$; as noted in this MathOverflow post, it is possible to do this. Let $\phi$ be an automorphism that is not inner. This automorphism will then have no nontrivial fixed points, and because $3\nmid p-1$, if $x$ is a nontrivial element of $T_p$, then $\langle x,\phi(x)\rangle=T_p$.
Let $L$ be the subgroup of $\mathrm{Aut}(T_p)$ generated by $T_p$ (viewed as inner automorphisms) and $\phi$. Then $T_p$ is normal and has index $3$ in $L$, with quotient of order $3$ generated by the image of $\phi$.
Let $H=T_p$. If $f\colon L\to H$, then $f$ is either trivial on $T_p$, or it restricts to an automorphism of $T_p$, since $T_p$ is simple and is generated by any two elements that do not commute. If it is trivial on $T_p$, then the image of $f$ is finite. If it is one-to-one of $T_p$, note that because $\phi$ has no fixed points then $\phi$ cannot map to the identity. Assume $f(\phi) = x\neq e$. Let $y$ be such that $f(y)=x$. Since the images of $y$ and $\phi$ under $f$ commute, we have that $$f(\phi(y)) = f({}^{\phi}y) = f(\phi)f(y)f(\phi)^{-1} = xxx^{-1}=x=f(y).$$ But $\langle y,\phi(y)\rangle = T_p$, which gives a contradiction, since $f(T_p)=T_p$, but $\langle f(y),f(\phi(y))\rangle=\langle x\rangle$ is cyclic of prime order. Therefore, $f$ cannot be one-to-one on $T_p$, and thus the image of $f$ is finite.
That means that $T_p$ only admits finite images from $L$. However, $L$ is an extension of $T_p$ by a finite group, so taking $G=L$, we have that the identity morphism $L\to G$ does not have finite image.
Thus, we have a group $L$, a group $H$ that admits only finite images from $L$, a subgroup $G$ such that $[G:H]\lt\infty$, but $G$ admits infinite images from $L$.