The secret of the PDA to CFG proof is recursion. CFG's are by definition good in recursion.
Sipser explicitly constructs a CGF and the Lemma's are provided to show his construction works. His CFG has nonterminals of the form $A_{pq}$ and the intuitive meaning is that they correspond to "computations of the PDA starting in state $p$ ending in state $q$ where at the start and at the end the stack is empty". The proof shows that the CFG generates strings on nonterminal $A_{pq}$ when the PDA reads the same strings on computations of the $A_{pq}$ type.
Here are the productions that should make this work.
(1) Choose any two matching instructions as follows. One instruction from state $p$ to state $r$ which pushes a stack symbol $u$ and reading $a$ from input, and one from state $s$ to state $q$ popping same $u$ reading $b$. Then $A_{pq} \to a A_{rs} b$.
Any computation of a PDA that pushes $u$ must later pop that symbol where the stack below has not been touched. That is recursion: the computation $A_{pq}$ is replaced by the computation $A_{rs}$ which is inside the original computation, but has now additional $u$ on stack, from state $r$. We compute until $u$ is popped, and this is possible in state $s$.
This is for specific $p,q,r,s$ that have matching push and pop instructions.
(2) The computation from $p$ to $q$ from empty stack to empty stack might have an empty stack on the way. Then we have to introduce an intermediate state $r$, more or less in order to continue from there. (We push and pop our first stack symbol, and now need to continue with the next stack symbol.) Thus $A_{pq}\to A_{pr}A_{rs}$.
This is for all $p,q,r$ in order to continue a computation.
(3) The empty computation is also of the right form, so we have $A_{pp} \to \varepsilon$ as it generates nothing.
I like to note that Sipser has a somewhat nonstandard proof. Most books have a similar recursive construction, but also include the nonterminal that is put on the stack. Hence its nonterminals are of the form $[p,u,q]$ meaning a computation from state $p$ to state $q$ with only $u$ on the stack until that $u$ is popped. Same intuition, technically different as Sipser "predicts" the final $q$ (it is chosen as a "pop" to match the initial "push") whereas many other books "guess" the state $q$ where the pop will occur.