How should we interpret a transition rule $\epsilon,b\to c$ in a Pushdown Automata as Sipser defined in his book? After defining pushdown automaton in Sipser's book at p. 113:



at the bottom of p.114, he tries to describe a way to make a diagram for pushdown automaton as following:



My question is about the part "If $a$ is $\epsilon$, the machine may make this transition
without reading any symbol from the input." But in many examples in the book, it seems some $\epsilon$s which are in place of $a$ must be interpreted as an empty string to show that the pointer is reached to the end of input. For example in the following diagram, we must interpret the first $\epsilon$ in the label of the transition between $q_3$ and $q_4$ to an empty string which shows the end of input string:



Otherwise, it can accept $01^2$. Since, if we interpret $\epsilon,\$\to\epsilon$ as to make this transition from $q_3$ to $q_4$ without reading the second 1, then $01^2$ would be accepted.

Now my question is how a PDA decides if that $\epsilon$ is for moving without reading any symbol from input or that $\epsilon$ is an empty string?
Suppose that your PDA is reading $011$.

At the beginning, the PDA is at state $q_1$, and the stack is empty. The only move the PDA can make is to put $\$$ on the stack and to transition to state $q_2$.

Since the next symbol to be read is $0$, the only move the PDA can make now is the self-loop on $q_2$ which puts $0$ on the stack. The input is now $11$, and the stack is $0\$$ (top of stack first).

Since the next symbol to be read is $1$, the only move the PDA can make is to transition to $q_3$, erasing $0$ from the top of the stack. The input is now $1$, and the stack is $\$$.

Since the next symbol to be read is $1$, the PDA must move to $q_4$, erasing $\$$ from the stack. The input is still $1$, and the stack is empty.

At this point, the PDA is stuck. It cannot make any move. However, it hasn't read all the input yet. We conclude that there is no accepting computation for this input, and so the PDA doesn't accept $011$.