Let $(M,g)$ be a Riemannian manifold and $\tilde{M}$ its universal cover. Suppose that $\pi_1(M)$ is finitely generated. Let $\tilde{g}$ be the lift of $g$ to $\tilde{M}$, and let $d$ denote the induced distance function on $\tilde{M}$.
Let $L\colon\pi_1(M)\to\mathbb{N}$ be a fixed length function, and let $F\subseteq\tilde{M}$ be a fundamental domain for the action of $\pi_1(M)$ on $\tilde M$.
(By fundamental domain, I mean that $F$ is the closure of an open subset of $\tilde{M}$ such that for any $g\in\pi_1(M)$, the interiors of $g\cdot F$ and $F$ have no common points, and $\{g\cdot F\}_{g\in\pi_1(M)}$ form a cover of $\tilde M$.)
Question: Given any sequence $(g_i)_{i\in\mathbb{N}}\subseteq\pi_1(M)$ such that $\lim_{i\to\infty}l(g_i)=\infty$, is it true that $$\lim_{i\to\infty}d(F,g_i\cdot F)=\infty,$$ where $g_i\cdot F$ is the translate of $F$ by the action of $g_i$?
Remark: If $M$ is compact, then the answer should be yes, since the action of $\pi_1(M)$ on $\tilde{M}$ is always proper. In the non-compact case, I would expect the answer to be no, but would like an example that shows this.
In my answer, I will assume that $$ d(A,B)=\inf_{a\in A, b\in B} d(a,b). $$
Here is a simple and classical example. Consider the modular group $\Gamma=PSL(2, {\mathbb Z})$ and its standard fundamental domain $F$. Let $u_n\in \Gamma$ be any sequence corresponding to (strictly) upper-triangular matrices. Then for each $u_n$, $d(u_nF,F)=0$.
Or, for a simpler example consider the subgroup $\Gamma< PSL(2, {\mathbb R})$ generated by the translation $u: z\mapsto z+1$ (in the upper half-plane model). As $F$ take $$ \{z: Im(z)>0, 0\le Re(z)\le 1\}. $$ Then for each $\gamma\in \Gamma$, $$ d(F, \gamma F)=0. $$