No. Let $X$ be $\mathbb R$. In the ring $C^{\infty}(X)$ let $I$ be the ideal of all functions vanishing to infinite order at $0$. The module $C^{\infty}(X)/I$ does not have a finite resolution by finitely generated projective modules.
Edit:
Still no if you want the finitely generated module to be contained in a finitely generated projective module. For the same $X$ pick a function $f$ such $f(x)$ vanishes precisely when $x<0$. let $J$ be the ideal generated by $f$. The module $J$ does not have a finite resolution by finitely generated projective modules.
For both of these examples, the method I have in mind is this: If a module $M$ has a finite projective resolution $P_\bullet$ then for every point in $p\in X$ the alternating sum of the $k_p$ vector space dimension of $Tor_n(M,k_p)$ is independent of $p$ because it's the alternating sum of the rank of $P_n$. I believe that in the first example this Euler number comes out to be $1$ if $p$ is the origin and otherwise $0$, and in the second it's $1$ if $p> 0$ and $0$ if $p< 0$.