Why no DPDA can accept Palindrome? (according to this proof)
No sequence of moves may empty the stack, this is seen as follows. Since the PDA is deterministic, its computation is determined by its input. So we have to argue that the computation on $x$ will not empty the stack. If it would empty the stack, then the PDA blocks, and can no longer continue. But observe that $xx^R$ is a palindrome, so there should be an accepting computation on $xx^R$. This implies that the computation on the prefix $x$ should not block.

Your second question. For input $x$ we can look at all continuations $xy$. For each $y$ determine the stack height that is obtained. We choose an $y$ for which this is minimal and call this specific string $y_x$ (to indicate it depends on $x$).