Of course, the correct definition of coherence is that in your Question 2. It just so happens that for a sheaf of modules on a scheme it is equivalent to the easier one.
As far as a I know, the notion of coherence is mostly used when one has a sheaf of ring (often non-commutative), different from the structure sheaf. So, for example, the sheaf of D-modules on a smooth variety is not noetherian, but it is a coherent sheaf of rings; this is a very important fact.
There are schemes whose structure sheaf is not coherent, and those are a bit of a mess; for example, the locally finitely presented quasi-coherent sheaves do not form an abelian category. However, in most cases one is not usually bothered by them.
For example, in setting up a moduli problem, it is quite useful to consider non-noetherian base schemes, because the category of locally noetherian schemes has problems (for example, is not closed under fibered products). For example, if $X$ is a projective scheme over a field $k$, and $F$ is a coherent sheaf on $X$, one defined the Quot functor from schemes over $k$ to sets by sending each (possibly non-noetherian scheme) $k$-scheme $T$ into the set of finitely presented quotients of the pullback $F_T$ of $F$ to $T$ that are flat over $T$. Of course, when you actually prove something, one uses that fact that locally on $T$ any finitely presented sheaf comes from one defined a finitely generated $k$-algebra, and works with that, free to use all the results that hold in the noetherian context. Thus, in practice most of the time you don't need to do anything with non-noetherian schemes.
The property that every coherent sheaf admits a surjection from a coherent locally free sheaf is also known as the resolution property.
The theorem can be refined as follows:
Every noetherian, locally $\mathbb Q$-factorial scheme with affine diagonal (equiv. semi-semiseparated) has the resolution property (where the resolving vector bundles are made up from line bundles).
This is Proposition 1.3 of the following paper:
Brenner, Holger; Schröer, Stefan
Ample families, multihomogeneous spectra, and algebraization of formal schemes.
Pacific J. Math. 208 (2003), no. 2, 209--230.
You will find a detailed discussion of the resolution property in
Totaro, Burt.
The resolution property for schemes and stacks.
J. Reine Angew. Math. 577 (2004), 1--22
Totaro proves in Proposition 3.1. that every scheme (or algebraic stack with affine stabilizers) has affine diagonal if it satisfies the resolution property.
The converse is also true for smooth schemes:
Proposition 8.1 : Let $X$ be a smooth scheme of finite type over a field. Then the following are equivalent:
- $X$ has affine diagonal.
- X has the resolution property.
- The natural map $K_0^{naive} \to K_0$ is surjective.
Best Answer
Imposing that you can resolve by a length $2$ sequence of vector bundles is too strong. What you want is that there is some $N$ so that you can resolve by a length $N$ sequence of vector bundles. By Hilbert's syzygy theorem, this follows from requiring that the scheme be regular. (Specifically, if the scheme is regular of dimension $d$, then every coherent sheaf has a resolution by projectives of length $d+1$.)
Here is a simple example of what goes wrong on singular schemes. Let $X = \mathrm{Spec} \ A$ where $A$ is the ring $k[x,y,z]/(xz-y^2)$. Let $k$ be the $A$-module on which $x$, $y$ and $z$ act by $0$. I claim that $k$ has no finite free resolution. I will actually only show that $A$ has no graded finite free resolution. Proof: The hilbert series of $A$ is $(1-t^2)/(1-t)^3 = (1+t)/(1-t)^2$. So every graded free $A$-module has hilbert series of the form $p(t) (1+t)/(1-t)^2$ for some polynomial $p$; and the hilbert series of anything which has a finite resolution by such modules also has hilbert series of the form $p(t) (1+t)/(1-t)^2$. In particular, it must vanish at $t=-1$. But $k$ has hilbert series $1$, which does not.
There is, of course, a resolution of $k$ which is not finite. If I am not mistaken, it looks like
$$\cdots \to a[-4]^4 \to A[-3]^4 \to A[-2]^4 \to A[-1]^3 \to A \to k$$