1. Yes. By hypothesis, $\xi$ is a direct summand of a f.g. trivial vector bundle, so $S(\xi)$ is a direct summand of a f.g. free module.

2. Yes. A f.g. projective $R$-module is a direct summand of a f.g. free module, meaning it's obtained by taking fixed points of the action of some idempotent $e \in M_n(R)$ acting on some $R^n$. When $R = C(B, \mathbb{R})$ there is a corresponding idempotent acting on the trivial vector bundle $\mathbb{R}^n$, since the endomorphism ring of $\mathbb{R}$ (the vector bundle) is $R$. It remains to verify that taking fixed points of this idempotent produces a vector bundle. At least if $B$ is locally connected, this follows from the observation that the rank of the idempotent $e$ is $\text{tr}(e)$, which takes a discrete set of values (from $0$ to $n$) and hence is locally constant. I think it ought to be true in general but I'm less confident I can write down an argument off the top of my head.

1. Yes. By hypothesis, $\xi$ is a direct summand of a f.g. trivial vector bundle, so $S(\xi)$ is a direct summand of a f.g. free module.

2. Yes. A f.g. projective $R$-module is a direct summand of a f.g. free module, meaning it's obtained by taking fixed points of the action of some idempotent $e \in M_n(R)$ acting on some $R^n$. When $R = C(B, \mathbb{R})$ there is a corresponding idempotent acting on the trivial vector bundle $\mathbb{R}^n$, since the endomorphism ring of $\mathbb{R}$ (the vector bundle) is $R$. It remains to verify that taking fixed points of this idempotent produces a vector bundle. (Edit: Previously this section included the hypothesis that $B$ is locally connected, which is unnecessary.) The rank of the idempotent $e$, as a function on $B$, is $\text{tr}(e)$. This takes a discrete set of values, namely the integers from $0$ to $n$, and so its preimages break up $B$ into a finite disjoint union of clopen connected components. On each of these components $\text{tr}(e)$ is constant, and from here it should be straightforward to produce local trivializations.

1. Yes. By hypothesis, $\xi$ is a direct summand of a f.g. trivial vector bundle, so $S(\xi)$ is a direct summand of a f.g. free module.

2. Yes. A f.g. projective $R$-module is a direct summand of a f.g. free module, meaning it's obtained by taking fixed points of the action of some idempotent $e \in M_n(R)$ acting on some $R^n$. When $R = C(B, \mathbb{R})$ there is a corresponding idempotent acting on the trivial vector bundle $\mathbb{R}^n$, since the endomorphism ring of $\mathbb{R}$ (the vector bundle) is $R$. It remains to verify that taking fixed points of this idempotent produces a vector bundle. But this follows from the observation that the rank of the idempotent $e$ is $\text{tr}(e)$, which takes a discrete set of values (from $0$ to $n$) and hence is locally constant.

In fact we get, mostly through abstract nonsense (the least abstract nonsense part is the verification that the category of vector bundles is idempotent complete), an equivalence of categories between vector bundles over $B$ which are direct summands of f.g. trivial vector bundles and f.g. projective modules over $C(B, \mathbb{R})$, since both of these categories are Cauchy completions of the same enriched category, namely the one-object $\text{Ab}$-enriched category with endomorphism ring $C(B, \mathbb{R})$.

1. Yes. By hypothesis, $\xi$ is a direct summand of a f.g. trivial vector bundle, so $S(\xi)$ is a direct summand of a f.g. free module.

2. Yes. A f.g. projective $R$-module is a direct summand of a f.g. free module, meaning it's obtained by taking fixed points of the action of some idempotent $e \in M_n(R)$ acting on some $R^n$. When $R = C(B, \mathbb{R})$ there is a corresponding idempotent acting on the trivial vector bundle $\mathbb{R}^n$, since the endomorphism ring of $\mathbb{R}$ (the vector bundle) is $R$. It remains to verify that taking fixed points of this idempotent produces a vector bundle. At least if $B$ is locally connected, this follows from the observation that the rank of the idempotent $e$ is $\text{tr}(e)$, which takes a discrete set of values (from $0$ to $n$) and hence is locally constant. I think it ought to be true in general but I'm less confident I can write down an argument off the top of my head.