Well, first of all, do you understand that the decays to three pions is CP-odd and to two pions is CP-even? I mean why you need the K1 (CP-even) and K2 (CP-odd). If CP is conserved, then the K1 and K2 are the Klong and Kshort.
If CP was not violated, then that would be always the case. However, in experiment we observed that the K2 (long) would decay into 2 pions (Fitch and Cronin)! You do that by exploiting that the Kshort lives for a shorter time; so if you have a beam of neutral Kaons and allow it to travel some distance you'll be dominantly left with K-longs. CP-violation.
Then the explanation comes from 2 components:
1. CP-indirect violation. As the name suggests you introduce CP-violation by doing something indirectly... in which case the Klong and Kshort are not exactly the CP-eigenstates, but a mixture of those)... So the Klong has both a component of K1 and K2 which would allow it to decay into 2 or 3 pions with probabilities depending on the CP-state mixture. This is parametrized by the parameter ##\epsilon## which is the mixing phase:
[itex]K_L = \frac{1}{\sqrt{1-|\epsilon|^2}} ( K_2 + \epsilon K_1 )[/itex]
So the ##K_L## decays to 3 pions via the ##K_2## and to 2 pions via the ##K_1## (which is weighted by the ##\epsilon##)
2. CP-direct violation, again as the name suggests, here you brute-heartedly violate the symmetry. In which case you have the direct violation of CP : ##K_L = K_2 \rightarrow \pi \pi## which is parametrized by the parameter ##\epsilon'##.
Both mechanisms contribute in the Kaon CP-Violation, but the 2nd not as strongly...
In fact you can go further into this and start thinking of the actual physics/mechanisms in those cases, so you would start breaking the phases into real parts (eg. the ##Re(\epsilon)##, the ##Re(\epsilon ')## and the imaginary parts: ##Im(\epsilon) , Im(\epsilon ')## measure different things in CP-Violation)