More specifically, I was wondering if there are wellknown conditions to put on $X$ in order to make $K_0(X)\simeq K^0(X)$. Wikipedia says they are the same if $X$ is smooth. It seems to me that you get a nice map from the coherent sheaves side to the vector bundle side (the hard direction in my opinion) if you impose some condition like "projective over a Noetherian ring". Is this enough? In other words, is the idea to impose enough conditions to be able to resolve a coherent sheaf, $M$, by two locally free ones $0\to \mathcal{F}\to\mathcal{G}\to M\to 0$?
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 $k[x,y,z]/(xzy^2)$. 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 $(1t^2)/(1t)^3 = (1+t)/(1t)^2$. So every graded free $A$module has hilbert series of the form $p(t) (1+t)/(1t)^2$ 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 $p(t) (1+t)/(1t)^2$. 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 $$\cdots \to a[4]^4 \to A[3]^4 \to A[2]^4 \to A[1]^3 \to A \to k$$ 


You want coherent sheaves to have finite global resolutions by locally free sheaves. So definitely you need the regularity of $X$ to ensure that a locally free resolution stops at a finite stage. You also need a global condition such as quasiprojectivity over an affine base to guarantee that you can start the process. (The last condition is not optimal.) Edit: In reading the follow up comments, I realize my answer was a bit cryptic. The inverse map $K_0(X)\to K^0(X)$ would send the class of a coherent sheaf to the alternating sum of the classes in a resolution. In general, these groups behave quite differently. $K^0(X)$ is contravariant like cohomology and $K_0(X)$ is covariant for proper maps like (BorelMoore) homology. That they coincide for regular schemes is reminiscent of Poincaré duality. 


Asking $K^0(X)$ to be isomorphic to $K_0(X)$ is not always "good enough". Of course, it will allow you to carry over constructions for $K_0(X)$ to $K^0(X)$, but not canonically. And it can happen that $K^0(X)\cong K_0(X)$ without $X$ being regular. For example, take $X= \textrm{Spec} A$, $A=k[x]/(x^n)$ with $n\geq 2$. Then you have an infinite resolution as given in David's answer for $k$. Computing $Tor^A_i(k,k)$ shows that $k$ has no finite resolution. (In fact, $Tor_i^A(k,k) = k^2$ for all $i>0$.) Now, although the above "existence of finite resolution" fails, it is not hard to see that $K^0(X)\cong \mathbf{Z}\cong K_0(X)$ in this case. (Use that $A$ is a local ring and the length map on $A$.) Of course, the natural map $K^0(X) \longrightarrow K_0(X)$ is not an isomorphism. (It is given by $1\mapsto n$.) [Edit: I added another example] [Edit 2: There was something wrong with the example below as noted by Michael. I fixed the problem] Let me also add to my answer the following "snake in the grass". If you work with general schemes, even if regular, one requires the extra assumption of "finitedimensionality". For example, take the scheme $X=\textrm{Spec} (k \times k[t_1]\times k[t_1,t_2] \times \ldots)$. Now, even though $A = k\times k[t_1]\times\ldots$ is regular, there is an infinite resolution for $k$ of the form $$\ldots \longrightarrow A\longrightarrow A\longrightarrow A \longrightarrow k \longrightarrow 0$$ which corresponds geometrically to taking a point, then adding a line, then adding a plane, etc. Again, take the Tor's to see that $k$ has no finite resolution. Do note that $X$ is not noetherian. [Edit 3: I added the following for completeness] Let $X$ be a regular finitedimensional scheme. Assume that $X$ has enough locally frees. (This notion also arose in Are schemes that "have enough locally frees" necessarily separated ). Then the canonical morphism $K^0(X) \longrightarrow K_0(X)$ is an isomorphism. In the second example, $X=\textrm{Spec} \ A$ is regular, but not finitedimensional. Does $X$ have enough locally frees? 

