Thinking of a f.g. projective module as a vector bundle, it seems very likely the answer is no: consider the trivial bundle of rank 2 and two "twisted" subbundles of rank 1 whose intersection is 0-dimensional everywhere except over a closed subset with non-empty interior – so not a vector bundle, in particular.
Let's see where this thinking leads. Let $R$ be the ring of real-valued continuous functions on the real line $\mathbb{R}$. Let $M = R \oplus R$. Let $f : \mathbb{R} \to \mathbb{R}$ be the following continuous function:
$$f (x) = \begin{cases}
\exp (-(x+1)^{-2}) & x < 1 \\
0 & -1 \le x \le 1 \\
\exp (-(x-1)^{-2}) & x > 1
\end{cases}$$
Consider the submodules below:
\begin{align}
N_1 & = \{ (s_1, s_2) \in M : f s_1 - s_2 = 0 \} &
N_2 & = \{ (s_1, s_2) \in M : f s_1 + s_2 = 0 \}
\end{align}
$M$, $N_1$ and $N_2$ are all clearly f.g. free $R$-modules. Yet their intersection is not f.g. projective:
$$N = N_1 \cap N_2 = \{ (s_1, s_2) \in M : f s_1 = 0, s_2 = 0 \}$$
Let $\mathfrak{m}$ be the maximal ideal of $R$ consisting of functions vanishing at $1$, and let $R_\mathfrak{m}$ be the localisation of $R$ at $\mathfrak{m}$. If $N$ were f.g. projective, then $N_\mathfrak{m}$ would be f.g. free over $R_\mathfrak{m}$. On the other hand, by construction, $N_\mathfrak{m}$ is annihilated by the germ of $f$, which is not zero, and $N_\mathfrak{m}$ is itself non-zero (take, for example, any bump function supported on $[-1, 1]$) – so $N_\mathfrak{m}$ cannot be free. Therefore $N$ is not f.g. projective.
Here is a picture of one of the monstrosities we constructed:
