$\delta(q_1,a,b) = \left\{(q_2, ba)\right\}$ means from state $q_1$ on input $a$ with stack top being $b$, the PDA moves to state $q_2$ and pushes $a$ on top of stack.
So, here the missing transitions are at the middle of the input string:
$\delta(q_1,a,a) = \left\{(q_2, \epsilon)\right\}$
$\delta(q_1,b,b) = \left\{(q_2, \epsilon)\right\}$
Once middle is reached, now we should start popping. And so, we must go to state $q_2$ as well as pop the previous character on the stack. (The character before and after the middle must be same as the string is even length palindrome)
(This is a non-deterministic PDA)