Your objection is correct, but the problem is easily fixed. Define $\dot\alpha$ to be $\{(\check\beta,a_\xi):\beta<\xi\}$. Then, because the $a_\xi$ form a maximal antichain, each $a_\xi$ will force $\dot\alpha=\check\xi$, and $\dot\alpha$ is forced (by all conditions) to be an ordinal. Furthermore, the name $\dot\alpha$ is in $N$ (because it's defined from the antichain $A$ and it enumeration, which are in $N$). By statement 2, $q$ forces $\dot\alpha\in\check N$. Some extension $r$ of $q$ forces a specific value, say $\xi\in N$, for $\dot\alpha$. So $r$ can't be compatible with $a_\eta$ for any $\eta\neq\xi$ (because $a_\eta$ forces $\dot\alpha$ to equal $\check\eta$). Therefore $r$ is an extension of $a_\xi$ (by maximality of the antichain and separativity of the forcing notion), which means $q$ is compatible with $a_\xi$. Finally, since both the enumeration of $A$ and the ordinal $\xi$ are in $N$, $a_\xi$ is also in $N$, and that completes the proof of statement 1.
A few minor comments: In case you don't require separativity as part of the definition of "notion of forcing", either pass to the separative quotient first, or note that even though my argument won't make $r$ an extension of $a_\xi$, it will make them compatible, and this suffices.
One of the comments entertained the notion that $N$ might be transitive. It won't be, except in trivial cases. Usually $\chi$ is large, and then every definable element of $H(\chi)$, for example $\omega_1$, will be in $N$, whereas a countable $N$ can't contain all the elements of $\omega_1$.
Since a structure and an elementary substructure must have the same language, the setup at the start of the question should have $(N,P,\in)$ in place of $(N,\in)$. By elementarity, $P$ will be a notion of forcing in $N$ iff it is one in $H(\chi)$, and that is equivalent to being a notion of forcing in $V$ if $\chi$ is big enough.