As Leo Moos already said in the comments, it is impossible to find a compactly supported solution $u$. For the sake of argument, consider the case $M = \mathbb{R}^n$ and $f$ is supported inside a ball $B$ and suppose such a function $u$ exists. Outside of $B$, the solution $u$ to $\Delta u = f$ is harmonic. If $u$ also has compact support then it follows from unique continuation that $u$ vanishes identically on the complement of $B$. Now on the ball $B$, we see that $u$ is the unique solution to the equation $\Delta u = f$ with $u$ vanishing on $\partial B$. This means our only hope is to take $u$ to be the solution to the problem on the ball $B$ with Dirichlet boundary conditions, extended by zero outside. This won't be smooth however, because the normal derivative for this solution is not zero.

In general, it is impossible to find a compactly supported solution $u$ to the equation $\Delta u = f$. For the sake of argument, consider the case $M = \mathbb{R}^n$ and $f$ is supported inside a ball $B$ and suppose such a function $u$ exists. Outside of $B$, the solution $u$ to $\Delta u = f$ is harmonic. If $u$ also has compact support then it follows from unique continuation that $u$ vanishes identically on the complement of $B$. Now on the ball $B$, we see that $u$ is the unique solution to the equation $\Delta u = f$ with $u$ vanishing on $\partial B$. This means our only hope is to take $u$ to be the solution to the problem on the ball $B$ with Dirichlet boundary conditions, extended by zero outside. In general, this won't be smooth however, or even differentiable, because the normal derivative for this solution is not necessarily zero.