What is the process transforming $K_L$ mesons into $K_S$ mesons? First of all, I could not find any good book/webpage where I can read about K mesons, can somebody point me to a good reading?
The second point is about $K_L$ and $K_S$ mesons. As I can read, they are eigenstates of $K^0$ with opposite strangeness. In principle they should decay into pions, but I was wondering if a $K_L$ can spontaneously change into a $K_S$ instead of decay into pions or there would be an interaction with protons/nuclei needed for this transformation to occur.
 A: *

*$K_L$s cannot spontaneously transform to $K_S$s in free space. They are eigenstates of the free hamiltonian and maintain their identity. Their mass difference is $\Delta m \sim 10^{-14} m$.  To effect a 
$K_L\to K_S$ transformation ("regeneration") they must go through matter and interact coherently and strongly with its nuclei: absolutely nothing spontaneous about it. I would strongly recommend changing your title, as the present one is odd.


Let's ignore CP violation to avoid mission creep and muddying the waters in the process, so just call 
$$  |K_{L}\rangle \sim \dfrac{1}{\sqrt2}\left( |K^{0}\rangle - |\bar{K^{0}}\rangle \right)  \\   |K_{S}\rangle \sim\dfrac{1}{\sqrt2}\left(|K^{0}\rangle + |\bar{K^{0}}\rangle \right).$$ 
The $|K^{0}\rangle ,  |\bar{K^{0}}\rangle $ states are production states, that is they have well-defined strangeness, but they do not propagate or decay. The $|K^{0}\rangle\sim \bar{s} d$ is the isopartner of $K^+$, with S=1; while  the  $|\bar{K^{0}}\rangle\sim \bar{d}s$ is the isopartner of $K^-$, with S=-1. 
They differ in strangeness by two, so only a doubly weak interaction mixes them, the celebrated box diagram. When this is  taken into account, the free eigenstates are $K_L, K_S$ with very different lifetimes, so a beam of neutral kaons will quickly purify to a beam of surviving $K_L$, which are longer lived. In vacuum, no conversion to $K_S$s. 
As this beam enters a nuclear target,
the two "strongly distinct"  states (since strangeness is preserved in the strong interactions) are treated very very differently. The $K^0$ scatters its quarks by the exchange of gluons, pions, whatever, and exchanges quarks with the nucleons and back: elastic & charge-exchange scattering. You may write 137000 Feynman diagrams. Coherently, and then goes out, slightly diminished.
The $\bar{K^{0}}$ is different. It may do all of the above, but it also has additional channels: Its s quark may replace a nucleon quark to form a hyperon, Λ or Σ, so it is absorbed and gone. Whatever the hyperon does (decay), it does not return to the original state, so the absorption crudely degrades the $\bar{K^{0}}$ beam more, and increases strangeness, if you must think of it that way. (People would frown, with justice). 
The crucial thing is that, since the two components of the $K_L$ wave function were absorbed and phase-shifted differently, many times (coherent scattering), the outgoing wave function is not $\dfrac{1}{\sqrt2}\left( |K^{0}\rangle - |\bar{K^{0}}\rangle \right) $ anymore, but a slightly distorted version thereof. When resolved to free propagation eigenstates, the distortion amounts to a component of $K_S$ now present: $K_S$ has been regenerated after strong interaction with matter.
The standard textbook by D. Perkins should answer any questions you might have.
