I'm reading the proof of Baer's criterion here:
https://ncatlab.org/nlab/show/Baer%27s+criterion
and I'm a little confused. Is it possible that they identify $M$ with its image in $N$?
I think if we are really formal, we should define a function $\tilde{f}: i(N) \to I: i(n) \mapsto f(n)$ and define the poset as
$$\{(M',f')\mid i(N) \leq M' \leq M, f': M' \to Q \quad morphism, f'\vert_{i(N)} = \tilde{f}\}$$
Applying Zorn's lemma yields a maximal element $(M',f')$ and it can then be shown that $M'=M$. Then for $n \in N$ we have
$$f'\circ i(n) = f'(i(n)) = \tilde{f}(i(n)) = f(n) $$
and thus $f' \circ i = f$, thus $f'$ is the extension of $f$ we are looking for.
Is the above correct? Did I unravel the identifications correctly?
Yes, you're right, you understood the matter of notation and the set to apply Zorn on is the right one. But observe that even if you did this proof by just considering an actual inclusion $N \subseteq M$ instead of a generic monomorphism $ i \colon N \rightarrowtail M$ then nothing would change.
Indeed, after having proved that, if $N \subseteq M$, then every arrow $N \to Q$ extends to an arrow $M \to Q$, consider an injection $N \xrightarrow{i} M$ and an arrow $N \xrightarrow{f} Q$. Then $(N \xrightarrow{i} M)=(N \xrightarrow{j} i(N) \subseteq M)$ and $j$ is an isomorphism. Hence the arrow $fj^{-1} \colon i(N) \to Q$ extends to an arrow $g \colon M \to Q$ along the inclusion $i(N) \subseteq M$, by what we proved before. But then the arrow $f \colon N \to Q$ extends to the arrow $g\colon M \to Q$ along the whole $i \colon N \to M$.
Hence you see that, if you prove the statement just for the actual inclusions, then the statement for every monomorphism follows for free.