How is the Ebit measurement (or Bell state measurement), if Charlie has an entangled state?

Assume Charlie has an entangled state like $psi_{00}|00rangle+psi_{11}|11rangle,$, and he wants to send one of the share qubits to Alice and then Alice does a Bell state measurement and then tells Bob the result of two bits, and then Bob recovers Charlie’s state. So if Charlie’s state is unentangled and for example one bit like $|0rangle$, then it is obvious. But my question is about the situation that Charlie’s state is entangled and then sends one share qubit to Alice. Now, I have some questions according to this circuit:

I should also mention that the Bell state of Alice and Bob’s qubits is $|beta_{00}rangle=(|00rangle+|11rangle)/sqrt2$.

1. How can I understand the step-by-step change of every single qubit in the below circuit?
2. Assume after recovering Charlie’s state by Bob, he sends it back to Charlie. So now, How can Charlie have his entangled state again (I really do not understand that)?
3. What happens if $psi_{imath imath} rightarrow 1$ for each $imath$ and $psi_{00}=psi_{11}$?

I should mention that I want also the resultant state for the input $psi_{00}|00rangle+psi_{11}|11rangle,$ with $psi_{imath imath}$ any complex numbers subject to the normalization condition, subject to any one of the four ebit measurements $left|b_{1} b_{2}rightrangleleftlangle b_{1} b_{2}right|$, which $b_1$ is the share qubit that Alice has and $b_2$ is the share qubit that Bob has.