Need for 3 rounds in Luby Rackoff theorem I get that using Feistel network we can create a secure PRP from a secure PRF (in 3 rounds according to theorem).

My question is why not just 2 rounds of Feistel network  is sufficient. Is it that the output is random enough compared to a truly random string before 3 rounds?
The simple answer is that fewer than 3 rounds can be easily distinguished from a random permutation. The 2-round Luby-Rackoff cipher on $2n$ bits, using random functions $f_i$ mapping $n$ bits to $n$ bits, consists of 
$$
F(L, R) = (A, B),
$$
where $A = L \oplus f_1(R)$ and $B = R \oplus f_2(L \oplus f_1(R))$. 

Now consider an attacker that wants to distinguish $F$ from a random permutation. First they send some arbitrary $(L_1, R_1)$ to the oracle, and get back $(A_1, B_1)$. 

Next they send $(L_2, R_1)$, which results in $(A_2, B_2)$. Verifying that $A_1 \oplus A_2$ = $L_1 \oplus f_1(R_1) \oplus L_2 \oplus f_1(R_1)$ = $L_1 \oplus L_2$, the attacker is now pretty sure this is the Luby-Rackoff cipher.

What happens at 3 rounds is that no such distinguisher (querying only in the forward direction, of course!) is now possible. The proof relies on showing that collisions on the internal variables $A_i, B_i$ are necessary to mount any distinguisher, and then shows that those collisions are rare, in that you need approximately $2^{n/2}$ oracle queries to obtain one.