A push down automation (pda) is given in the following extended notation of finite state diagram:

The nodes denote the states while the edges denote the moves of the pda. The edge labels are of the form $d$, $s/s'$ where $d$ is the input symbol read and $s, s'$ are the stack contents before and after the move. For example the edge labeled $1, s/1.s$ denotes the move from state $q_0$ to $q_0$ in which the input symbol $1$ is read and pushed to the stack.

Introduce two edges with appropriate labels in the above diagram so that the resulting pda accepts the language $\left\{x2x^{R} \mid x \in \left\{0,1\right\}^*,x^{R} \text{ denotes reverse of x}\right\}$, by empty stack.

Describe a non-deterministic pda with three states in the above notation that accept the language $\left\{0^{n}1^{m} \mid n \leq m \leq 2n\right\}$ by empty stack

A Thing to note here is s is anything or like any symbol on the top of the stack so whatever the top of the stack .S represents in that way not any special symbol of the language

Say for some word $0112110$ we have to push every thing into the stack till $2$ . then we get $1$ then $1$ will be at top of stack so pop it or if get $0$ then $0$ will at top of stack so pop it. For any word of language it is applicable. $2$ is a mark that tell now we have to pop $0$ for $0$ and $1$ for $1$.

@shraddha there should be one more transition on q2, eps| stackSymbol| eps, but seems it's some hypothetical PDA and they don't have any stack symbol at the bottom of the stack. so once input is a member of the language and when input is finished stack will be empty.

The no. of 0's in stack is between no. of 0's in string and its double (all possible due to non-determinism). Now, the string is accepted if the no. of 1's in string matches the no. of 0's on stack.

@saurav you have to remove the last 2 transitions. Otherwise strings like 00001 will be accepted.

can we do it like this: for each 0 push two zeroes.When a 1 comes pop 2 0s.If stack becomes empty and there are still 1 left keep on reading it or in other words do nothing just read them till input gets over.Or if $ is reached and top of stack is s then also accept it.

δ(q_{1},∊, z_{0}) |---- (q_{1},∊) . This transition is required, right? otherwise, how stack top symbol will be popped? And if it is not popped then how stack will be emptied?

$\lambda$ in the stack part is used to indicate "whatever be the input". And the additional $(\lambda,Z,\lambda)$ was added to pop the initial symbol from the stack.