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.
The way I think about this (and it should be isomorphic to the way any reputable source does) is the following:
$L$ is locally equal to $\operatorname{Spec} S(\mathcal{F})$, the spectrum of the symmetric tensor algebra over $\mathcal{F}$. This is pretty easy to understand on an affine patch—you are taking formal sums of tensors of module elements $a+b\otimes c + d\otimes e\otimes f \otimes g + u\otimes w \ldots$, where you quotient out by $x\otimes y - y\otimes x$ so that it's commutative.
Since $\mathcal{F}$ is locally free, it's not too hard to see that $S(\mathcal{F})$ is locally isomorphic—over an open affine patch $\operatorname{Spec} A=U\subset Y$ small enough to trivialize $\mathcal{F}$—to $A[x_1,\ldots,x_n]$, where $n$ is the rank of $\mathcal{F}$.
Let me make that explicit. On $U$, $\mathcal{F}$ looks like the module $M=\bigoplus_{i=1}^n Ax_i$. Then $S(M)$ is exactly the free algebra over $A$ generated by $x_1, \ldots , x_n$. So $S(M) = A[x_1,\ldots,x_n]$, and we can see that it's reasonable to call $\operatorname{Spec} S(\mathcal{F})$ a "vector bundle", since locally it looks like $\mathbb{A}^n_U$.
Okay, now let's suppose that $X\to Y$ is locally given by a morphism of rings $A\to B$, where $\operatorname{Spec} B = V \subset X$. The pullback of $M$ is clearly the base change $M\otimes_A B = \bigoplus_{i=1}^n B x_i$.
But how about pulling back the bundle? Well, it's just $\mathbb{A}^n_U \times_U V = \mathbb{A}^n_V$. Or, on the level of rings, we have $S(M) \otimes_A B = S(M\otimes_A B)$. And that works at the sheaf level, too: $S(\mathcal{F}) \otimes_{\varphi^{-1} \mathcal{O}_Y} \mathcal{O}_X = S(\mathcal{F}\otimes_{\varphi^{-1} \mathcal{O}_Y} \mathcal{O}_X)$.
So this all adds up to a pretty strong dictionary (actually, an equivalence of categories) between locally free sheaves and so-called "geometric vector bundles". Actually, it's a little more general than that—if you look closely at how morphisms of sheaves lift up to the corresponding bundles, you find that as long as $\mathcal{F}$ is reflexive—that is, the natural map $\mathcal{F}\to\mathcal{F}^{\vee\vee}$ is an isomorphism—it can pretty much be identified with its bundle—the sheafy $\operatorname{Spec} S(\mathcal{F})$. You can think of these as singular bundles.
Best Answer
The idea is to interpret the sheaf of section of the bundle $E$ as the locally free sheaf you mean. Since $E$ is a vector bundle, over small enough $U \subset Y$, $E|_U \simeq U \times k^n$, and the sections of the bundle $U \times k^n \to U$ correspond exactly to maps $U \to k^n$, whence we get isomorphism with $\mathcal{O}_U^n$.
Turning locally free sheaf $\mathcal{E}$ into a vector bundle $E \to Y$ is similarly simple: if our sheaf is free over each of $U_i$, and $\bigcup U_i = Y$, then we just need to glue $U_i \times k^n \to U_i$ over various $U_i$. Therefore, we need a family of gluing isomorphisms $u_{ij}: (U_i \cap U_j) \times k^n \to (U_j \cap U_i) \times k^n$.
Here's how we get them: we have local isomorphism $f_i: \mathcal{E}_{U_i} \to \mathcal{O}_{U_i}^n$ from the definition of "locally free". Restricting to intersection $U_i \cap U_j$, we get $f_{ij} = f_j|_{U_i \cap U_j} \circ f_i|_{U_i \cap U_j}^{-1}: \mathcal{O}_{U_i}^n|_{U_i \cap U_j} \to \mathcal{O}_{U_j}^n|_{U_i \cap U_j}$. Every such map is induced by a map $u_{ij}: (U_i \cap U_j) \times k^n \to (U_j \cap U_i) \times k^n$ (why?), and this is our gluing isomorphism.