It is well-known that if a complete Riemannian manifold has bounded curvature and injectivity radius bounded away from zero, then the space $C^\infty_c(M)$ is dense in the Sobolev spaces $W^{k, p}(M)$ for $k=0, 1, 2$.
My question: Is this an if and only if? That is, if $C^\infty_c(M)$ is dense in these Sobolev spaces, does $M$ necessarily have bounded curvature and injectivity radius bounded away from zero?
The general answer to this question is no. Global bound on the Ricci curvature is not necessary for the density of smooth functions with compact supports.
Indeed, when $(M,g)$ is a smooth complete Riemannian manifold with positive injectivity radius and lower bound for the Ricci curvature, then the smooth functions with compact support are density in the Sobolev spaces for $p$ equals 2. This can be found for instance on Emmanuel Hebey's book: Nonlinear Analysis on Manifolds: Sobolev Spaces and Inequalities.
I'd say it is always dense.
Certainly, for any point $x\in M$ there is a nbd $U$ such that any $f\in W^{p,k}(M)$ supported in $U$ can be approximated by smooth functions supported in $U$ (e.g. for the same result you are quoting: we could modify $M$ in the complement of $U$ to make it compact, and nobody that lives inside $U$ would realize it).
Smooth partitions of unity do exist in any Riemannian manifold $M$. Therefore, any $f\in W^{p,k}(M)$ can be written as a locally finite sum $f=\sum_n f_n$ with $f_n\in W_c^{p,k}(U_n)$, for some locally finite open cover $\{U_n\}_n$ of $M$, by small nbd's $U_n$ were the density is true. Given $\epsilon>0$, we can choose $g_n\in C^\infty_c(U_n)$ such that $\|f_n-g_n\|_{p,k}\le 2^{-n}\epsilon$, so that the corresponding $g=\sum_n g_n$ is a smooth function with $\|f-g\|_{p,k}\le\epsilon$.
