Are schemes that "have enough locally frees" necessarily separated

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:

1. $X$ has affine diagonal.
2. X has the resolution property.
3. The natural map $K_0^{naive} \to K_0$ is surjective.

Actually the implication should be reversed: a separated regular Noatherian scheme has enough locally frees (this is Exercise 6.8, Chapter III Hartshorne). So the hypothesis is certainly needed for the proof.

EDIT: The statement in Hartshorne assumes X is integral, but this is not needed: see SGA 6, 2.2.3, 2.2.4, 2.2.5 and 2.2.7.1 (page 168-172 here ). In particular, you need separatedness and locally factorial (which follows from regular) to show that any coherent sheaf is a quotient of a direct sum of line bundles (for precise statement and example see 2.2.6 and 2.2.6.1).

In summary, the statement of the theorem is : For $X$ a regular, Noetherian, separated scheme one has $K_0(X) \cong K^0(X)$.

The answer to your question is no, as pointed out by Antoine.

This doesn't answer your question but it is may be worth noticing that any vector bundle $E$ on the affine plane $X$ with doubled origine is trivial. Indeed, the inverse image of $E$ via the two natural maps $u_i\colon A^2\rightarrow X$ are vector bundles on $A^2$, so are trivial. The glueing condition on $X\setminus\{o_1,o_2\}$ ($o_1,o_2$ are the two origins) is an automorphism of the trivial line bundle on $A^2\setminus\{o\}$, hence extends to an automorphism on $A^2$ by Hartogs. This implies that the initial vector bundle is trivial.

By the way, the affine line with doubled origin certainly has enough locally frees...