First, I want to make sure I understand why $\omega^M$ is an initial segment of $\omega^N$: since $M$ thinks of itself as standard, and of $N$ as non-standard, $M\vDash\forall n<\omega (n\in\omega^N)$, so by looking from the outside we get $\forall n<\omega^M (n\in\omega^N)$. Is that right?
That's exactly right.
Second, and more importantly, I don't understand why it follows that they have the same standard system. Is it something immediate that I miss, or a well-known result?
Let $n\in \omega^M$ be nonstandard. Then for any $a\in M$, $a\cap \omega=(a\cap n)\cap\omega$, so to determine the standard system of $M$, we can restrict to just subsets of $n$. But $M$ thinks $n$ is finite, so $M$ thinks $N$ must contain all subsets of $n$. So the entire standard system of $M$ must be in the standard system of $N$. Conversely, if $a$ is an element of $N$, then $a\cap\omega^M\in M$ and this has same intersection with the true $\omega$ as $a$, so the standard system of $N$ is contained in the standard system of $M$.
(Here I am abusing notation slightly and identifying $\omega$ with the canonical initial segment of $\omega^M$ which is isomorphic to it and $\omega^M$ with the canonical initial segment of $\omega^N$ which is isomorphic to it. Since the element relation of $M$ is not actually $\in$ and the element relation of $N$ is not actually the element relation of $M$, things like "$a\cap \omega$" or "$a\cap\omega^M$" which I wrote don't make sense literally.)
Actually, more strongly, you can prove that if $M$ and $N$ are arbitrary $\omega$-nonstandard models of ZFC (so we're not assuming $N\in M$) and $\omega^M$ is isomorphic to an initial segment of $\omega^N$ over the language of arithmetic, then $M$ and $N$ have the same standard systems. To prove this, let $m\in\omega^M$ be any nonstandard element. Note that for any $a\in M$, $M$ thinks $a\cap m$ is a finite set. It follows that $M$ can encode the set $a\cap m$ with an element of $\omega^M$, by the usual method of encoding finite sets of natural numbers in arithmetic. Identifying $\omega^M$ with its isomorphic initial segment of $\omega^N$, this element will also encode the set $a\cap m$ in $\omega^N$, so $a\cap m$ must also be a set in $N$. It follows that the standard system of $M$ is contained in the standard system of $N$. The converse can be proved the same way, since the natural number encoding a subset of $m$ in $N$ will still be an element of the initial segment $\omega^M$.