Let me address the modified question (as corrected by Jason). There are many intuitions depending on which proof-style you want to use. Here are a few remarks. 

The original  <a href="http://www.pnas.org/content/39/12/1268.full.pdf?sid=963a43be-495a-497f-981f-619a39300b03">proof</a> of Kodaira uses the curvature properties directly to show that there are no nonzero harmonic forms representing elements of $H^q(X, K_X\otimes L)$, when $q>0$ and $L$ is positive. In some sense, this is an adaptation of an argument of Bochner that shows that certain Betti numbers vanish for positively curved Riemannian manifolds (like the sphere). There is a slightly slicker proof -- again using harmonic forms -- of a somewhat stronger result due to Akizuki and Nakano. This is the one you find in many textbooks such as Griffiths and Harris. I think this is quite readable, but too long to explain here. One consequence of Kodaira's vanishing, as originally stated, is to prove that a line bundle is positive iff it is ample. Most algebraic geometers prefer to make this substitution in the statement, and I will do it as well below.

The Akizuki-Kodaira-Nakano vanishing implies the Lefschetz hyperplane theorem. Ramanujam turned this around to show that Lefschetz theorem together with the Hodge decomposition actually implies AKN-vanising. There are a bunch of other "topological" proofs to due Esnault-Viehweg and Kollár. Personally, I find these more insightful than the original arguments, but it is really a matter of taste.



There is now a purely algebraic proof due to Deligne-Illusie-Raynaud. 
The idea is to use a boot-strapping argument. If you can show that
$\dim H^q(X, K\otimes L)\le \dim H^q(X, K\otimes L^N)$ for $N\gg 0$. You get Kodaira from Serre vanishing. In order to actually get such an inequality, you have to use certain tricks with the Frobenius (i.e. you work in characteristic $p>0$). For example if it happens   that after reducing mod $p$, the map $O_X\to F_*O_X$ splits, then the above inequality holds for all $N=p^n$ by the projection formula. Unfortunately, this strategy usually fails for general $X$, but Deligne and Illusie use a more subtle splitting argument which works for all smooth $X$.