A standard proof for the fact that any 2 saturated models of the same cardinality are isomorphic can be found here
But I have doubt about this. Specifically, the construction of the next partial elementary map for successor ordinal. The problem here is at "$f_{\alpha}=f_{\beta}\bigcup\{(n_{\beta},b),(c,m_{\beta})\}$". Here we are extending the domain and range of the function. But there is nothing here that guaranteed that we still have a function at the end. $f_{\beta}$ might have already mapped $c$ to somewhere else, and now that $f_{\alpha}$ maps $c$ to $m_{\beta}$, $f_{\alpha}$ is no longer a function.
I don't see any ways to fix this problem. Anyone help?
You choose $c$ to realise the type $q(v) = \{\phi(v, \bar a) : \bar a \in A_\beta \cup \{n_\beta\} \text { and } \mathcal M \models \phi(m_\beta, f_\alpha'(\bar a))\}$. Now if $c \in A_\beta \cup \{n_\beta\}$, then the formula $v = c \in q$. Hence $\mathcal M \models m_\beta = f_\alpha'(c)$. So the new value of $c$ agrees with the old one.