What is sheaf cohomology intuitively? 
For local systems it is ordinary cohomology with twisted coefficients. But what 
if the sheaf in question is far from being constant? 
Can one still understand sheaf cohomology in some "geometric" way?
For example I would be very interested in the case of coherent $\mathcal{O}\_X$-Modules. 
Or even just line bundles.



What is sheaf cohomology intuitively?
One way to think about $H^1(A)$ is to use the long exact sequence not as a property of cohomology, but outright as a definition. That is, given an exact sequence of sheaves,
$$ 0\rightarrow A\rightarrow B\rightarrow C\rightarrow 0$$
then $H^1(A)$ is measuring the obstruction of global sections to be exact:
$$ 0\rightarrow \Gamma(A)\rightarrow \Gamma(B)\rightarrow \Gamma(C)\rightarrow H^1(A)$$
In words, $H^1$ is `measuring the failure of $\Gamma$ to preserve surjectivity'.
If you want this idea to actually *define* $H^1(A)$, you have to be careful to chose $B$ so that $H^1(B)=0$. But, as far as intuition goes, this works pretty well for me.


A demonstrative example of this, at least for me, is to think about the the complex variety $\mathbb{P}^1$, with $\mathcal{O}$ the structure sheaf and $\mathbb{C}\_p$ the skyscraper sheaf at a point. Then there is a surjective map (it is surjective because it is surjective on stalks):
$$ \mathcal{O}\rightarrow\mathbb{C}\_p$$
which has kernel $\mathcal{O}(-1)$, the twisted structure sheaf. This whole short exact sequence can be twisted by $(-1)$, noting that twisting a skyscraper sheaf $\mathbb{C}\_p$ gives an isomorphic sheaf $\mathbb{C}\_p(-1)$ (which I identify with the original sheaf):
$$0\rightarrow \mathcal{O}(-2)\rightarrow \mathcal{O}(-1)\rightarrow \mathbb{C}\_p\rightarrow 0$$


On global sections, we then get
$$ 0\rightarrow \Gamma(\mathcal{O}(-2))\rightarrow \Gamma(\mathcal{O}(-1))\rightarrow \Gamma(\mathbb{C}\_p)\rightarrow H^1(\mathcal{O}(-2))$$
We know that $\Gamma(\mathcal{O}(-1))=0$ and $\Gamma(\mathbb{C}\_p)=\mathbb{C}$, so the middle arrow is no longer surjective. Hence, $H^1(\mathcal{O}(-2))$ must contain at least $\mathbb{C}$ (in fact, it is exactly $\mathbb{C}$, since $H^1(\mathcal{O}(-1))=0$).


Higher cohomology may be also thought of this way: $H^{i+1}$ measures the failure of $H^i$ to preserve surjective maps. However, I don't find this very useful for thinking about higher cohomology, since it would need that I somehow understand lower cohomology much better.


