Let $M$ be a manifold and $H^k(M)$ denote its various Sobolev spaces of functions.
If $M$ is compact, then the Rellich lemma states that multiplication by the constant function $1$ gives a compact operator $H^k(M)\rightarrow H^l(M)$ whenever $k>l$.
Can one use this to show that, when $M$ is non-compact, the operator given by multiplication by a compactly supported function $f\in C_c(M)$ is compact $H^k(M)\rightarrow H^l(M)$ for $k>l$?
Yes, it works like that. Let $N$ be a compact manifold (maybe with boundary, the boundary does not matter when functions are compactly supported) that contains the support of $f$. You have a bounded multiplication operator $u\mapsto fu$ from $H^k(M)$ to $H^k(N)$, followed by the compact embedding $H^k(N)\to H^l(N)$. So the composition is compact.
If the presence of manifold-with-boundary is a concern, the problem can even be reduced to Euclidean space. Take a finite partition of unity $\{\varphi_k\}$ covering the support of $f$, so that each $\varphi_k$ is supported on a coordinate patch. Consider a bounded sequence $(u_j)$ in $H^k(M)$. For each $k$, the sequence $(f\varphi_k u_j)$ has a convergent subsequence in $H^l$ norm, by the Euclidean case of Rellich lemma. So we can pick a subsequence $u_{j_m}$ such that $(f\varphi_k u_{j_m})$ converges in $H^l$ norm for every $k$, hence $(f u_{j_m})$ converges.