When is the sheaf corresponding to a vector bundle on a smooth manifold coherent?

In algebraic and analytic geometry, vector bundles are usually interpreted as locally free sheaves of modules (over the structure sheaves). They are in particular examples of quasi-coherent sheaves. If the bundle is of finite rank, then the sheaf is actually coherent and this is good for certain cohomology groups to have finite rank for example.

I think the equivalence of vector bundles and locally free sheaves holds as well for the categories of topological spaces and smooth manifolds, and locally free sheaves are in particular quasi-coherent. The question is, when are they coherent? maybe it is best to ask first about the structure sheaf itself. In algebraic geometry the structure sheaf is coherent for neotherian schemes and in analytic geometry the structure sheaf of a complex manifold is coherent. What about smooth manifolds for example?

Since coherence is a local property and is preserved by finite direct sums, the general question reduces to the structure sheaf. But this is coherent only in trivial cases.

If $M$ is a smooth manifold whose connected components have positive dimension, then $C^{\infty}_M$ is not coherent.

For a proof, see Prop. 7.3.8 in the course notes by Andrew Lewis on sheaf theory. By a projection argument, it suffices to deal with $M=\mathbb{R}$. If $f$ is a smooth function, which is $>0$ on $\mathbb{R}_{>0}$ and $=0$ on $\mathbb{R}_{\leq 0}$, then the kernel of the multiplication map $f : C^{\infty}_{\mathbb{R}} \to C^{\infty}_{\mathbb{R}}$ is not of finite type: Loot at the stalk at $0$. The kernel $I$ is given by those germs of smooth functions vanishing on $\mathbb{R}_{\geq 0}$. If $\mathfrak{m}$ is the maximal ideal of functions vanishing at $0$, then we have $I = \mathfrak{m} I$ by Taylor expansion. Since $I \neq 0$, Nakayama's Lemma shows that $I$ is not finitely generated.