Why the long lived Kaon can not decay into two pions? The short-lived and long-lived states of kaon $|K_1>$ and $|K_2>$ respectively have the following compositions if they are the eigen states of CP parity:
$|K_1> = \frac{|K^0>\:-\:|\bar{K^0}>}{\sqrt2}$
$|K_2> = \frac{|K^0>\:+\:|\bar{K^0}>}{\sqrt2}$
In the book "Introduction to elementary particles" by David Griffiths, in the section 'Symmetries' for Neutral kaons it is given that $|K_1>$ can decay to two pions under CP symmetry and with right combination of orbital angular momenta for three pions system of $\pi^0 \pi^+ \pi^-$ which has CP parity = $+1$, $|K_1>$ can decay in to three pions. But $|K_2>$ can never be decay to two pions.
But wouldn't it be possible to use the same argument as three pions case for two pions with right combination of orbital angular momenta, which is antisymmetric under parity and is antisymmetric in isospin part be in total a symmetric wave function which gives CP parity $-1$. Then isn't it possible for $|K_2>$ to deacy to two pions under CP parity?
Experimentally it is observed that it's possible for 2 pion decay, that is the CP violation, but why theoretically it is not allowed?


 A: 
But wouldn't it be possible to use the same argument as three pions case for two pions with right combination of orbital angular momenta,

The argument is not about orbital angular momentum, but of the angular momentum any two particles have by definition of angular momentum as

The angular momentum of a particle of mass m with respect to a chosen origin is given by

$L=rXp$ where r is the distance from the chosen origin and p the momentum vector.
In the case of Kaon pi decays the origin is considered the Kaon, but there is no orbital angular momentum defined for particles that are not bound . The Pions coming out of the decay of Kaons are not bound to each other, only correlated by quantum numbers and conservation laws. So your use of ad hoc orbital angular momentum is not correct.
A: 
I am asking why it is not possible for two pions to have CP=-1? Griffiths says that part as a footnote to the mentioned section. I know I am missing something important, and that's why I am asking this. So are you saying two pions, when we consider their relative orbital angular momenta can not be antisymmetric? But for three pions it can be symmetric or antisymmetric?

What footnote? Be more specific; I suspect an impossible rabbit hole. Two pions cannot have  CP=-1. But as @anna v reminds you, both $_{1,2}$ have spin 0 and decay to an s-state, (L=0); so in symmetric combinations of pions.
Two such pions have P=+, and three have P= - . In both cases, you have C= + . So, CP=+ for 2 pions (the daughters of $K_1$, the short-lived θ); and - for 3 pions, (daughters of $K_2$, the long-lived τ). Griffiths hammers this in in (4.70), and the sentence right above it, "$K_1$ decays into 2π (never three); $K_2$ decays into 3π (never two)."
(But since you appear obsessed with this irrelevancy, other particles decay to 2 pions, as well, at higher spins, like p-wave (L=1): the ρ meson. It has P=-, C=-, hence CP=+ again! The parity switch and the charge interchange, e.g., for $\pi^+\pi^-$, cancel each other.  Go down the PDG listings and look for decays violating this systematic rule—in vain.)

Post OP-edit-footnote commentary
With your edit, that you should have included right away in your question, your puzzlement appears less preposterous.
Above, I'm quoting from the 1st edition, while you are quoting (and being needlessly confused by) the second, revised edition (2008). A truly unfortunate hedging footnote, as it tripped you into the rabbit hole. Still, it states expressly that the "same logic"  cannot apply, virtually by conservation of angular momentum, as already described above: there is no way to add orbital L to two spin 0 particles to get a spin 0 particle with negative CP: the two pions must be in an s-wave.
Nevertheless, for academic completeness, I can expand to shed light  on the logic of the ("actually...") defensive footnote, addressed to fussbudget  students, and trusting (?) it won't confuse them... Indeed, there is a very rare $\pi^+\pi^-\pi^0$ decay mode of $K_1\sim K_S$, with BR a million times smaller than the 2π mode, with mixed symmetry, so not a necessarily fully symmetric one like $3\pi^0$ which must violate CP! (We are assuming CP conservation throughout in this setup.) It is the smallest $K_s$ decay mode inferred and reported (CPLEAR at CERN), Sec 2.2.3. Note isospin is not conserved in this weak decay: the relevant part of the weak hamiltonian yields $\Delta I =3/2$ to reach an I=1 final state. (The mixed space symmetry is matched by the mixed isospin symmetry to amount to an overall Bose symmetry, as they should: Charged pion mutual interchange A, charged-neutral pion interchange  S!)
As we saw above, a $\rho^0 ~~(J^{PC}=1^{--})$  is thinkable as a p-wave (L=1) combination of $\pi^+ \pi^-$; presumably you have learned that it does not decay to $\pi^0\pi^0$ and why (antisymmetry). Think of combining this $\rho^0$ a with a $\pi^0~~ (J^{PC}=0^{-+})$ also in a p-wave, so with L=1, to a spin 0 state flipping P and resulting in a $J^{PC}=0^{--}$ extremely exotic state, so with PC=+, the conventional "nono" for three pions.
(But I am a little out of my depth here; the  $a_0(980)~~ (J^{PC}=0^{++})$ also has CP=+ and, being G-parity odd, can decay to 3π,  although I know little about its chiral lagrangian couplings. In any case, it would only figure in a footnote of a footnote of your footnote... If you are really, really curious, try the appendix of this unpublished preprint.)
