Let $G$ be free group with countable-infinitely many generators acting on $\Bbb R^2$ properly, i.e., for each point $x\in \Bbb R^2$ we have an open neighborhood $V_x$ of $x$ such that $g\cdot V_x\cap V_x=\varnothing$ for all non-trivial element $g\in G$.
Let $H\subsetneqq G$ be an infinite cyclic subgroup, so we have a covering map $p\colon \Bbb R^2\to \Bbb R^2/H$ as $H$ also acts properly on $\Bbb R^2$. Let $g\in G\backslash H$ and $\ell \colon\Bbb R\to \Bbb R^2$ be a proper map, i.e., the inverse image w.r.t. $\ell$ of a compact set is compact.
Is it true that the restriction of the covering map $p\colon \Bbb R^2\to \Bbb R^2/H$ on the subset $g\cdot \text{im}(\ell)$ of $\Bbb R^2$ is a proper map?
So, if $K\subseteq_\text{compact} \Bbb R^2/H$ is contained in an admissible open set of the covering $p$, then we have $p^{-1}(K)=\bigcup_{h\in H}h\cdot C$ for some compact set $C\subseteq_\text{compact} \Bbb R^2$. Now, one needs to show there are only finitely many $h\in H$ for which $hC\cap g\cdot \text{im}(\ell)\not=\varnothing$.
$\bullet$ My guess is that we need to also assume $G$ acts cocompactly on $\Bbb R^2$ i.e. there is a compact subset $S$ of $\Bbb R^2$ with $G\cdot S=\Bbb R^2$. But if I assume so, then by Lemma 1.17(b) $G$ has to be finitely generated.
What you ask for is not true.
For a counterexample, start with $\ell' : \mathbb R \to \mathbb R^2$ which is an "axis" for $h$, meaning a proper embedding which is invariant under $h$, such that the restriction of $h$ to $\text{im}(\ell')$ is a translation homeomorphism.
Notice: the map $p : \mathbb R^2 \to \mathbb R^2 / H$ restricted to $\text{im}(\ell')$ is not proper, because $p(\text{im}(\ell'))$ is a circle and hence is compact, but the inverse image of this circle under $p \bigm| \text{im}(\ell')$ is all of $\text{im}(\ell')$ which is homeomorphic to $\mathbb R$ and hence is not compact.
Now define $\ell = g^{-1} \circ \ell' : \mathbb R \to \mathbb R^2$. It follows that $p$ is not proper on $g \cdot \text{im}(\ell) = \text{im}(\ell')$.