Finitely generated projective modules are locally free

Yes, this is true. See this Math Overflow question for a precise statement and a reference to its proof in Bourbaki's Commutative Algebra.

This result is also stated in my commutative algebra notes, but the proof is not unfortunately not yet written up there. I certainly hope that this will be remedied soon though, as I will be teaching a course out of these notes starting on Monday. When the proof gets written, I will update this answer with a page number.

Added: Here is something in the MO answer that I decided was worth a comment here. For finitely generated modules, this stronger version of local freeness is actually equivalent to projectivity, whereas the weaker "pointwise local freeness" is subtly weaker in general.

For future reference, I have written up a constructive and reasonably self-contained, if somewhat dense, proof (one page). The basic idea is to first verify that idempotent matrices over local rings are equivalent to diagonal matrices with entries $1$ and $0$, thus showing that finitely generated projective modules over local rings are free.

This answer shows that if its stalk $M_\mathfrak{p}$ at $\mathfrak{p}$ is free, then there is an open neighbourhood $\mathfrak{p}\in D(f)$ on which its value $M_f$ is also free.

Proof: Reduce to the case that the natural map $M\to M_\mathfrak{p}$ is an injection by localising some $f_0\in A$. Now use the basis of $M_\mathfrak{p}$ to give a surjection $\alpha:A^n\to M$ whose localisation at $\mathfrak{p}$ is the isomorphism $A^n_\mathfrak{p}\to M_\mathfrak{p}$. Thus $\ker \alpha_\mathfrak{p}=0$, so $\ker \alpha_f=0$ for some $f\in A$. Thus $M_f\simeq A^n_f$ is free, completing the proof.

Thus the quasicompact space $\text{Spec}A$ admits a finite cover by $D(f_i)$'s, on each of which $M_{f_i}$ is free. $\text{Spec}A=\cup D(f_i)$ means precisely that $(f_1,...,f_n)=A$.