In particular, if $\mathcal{N}$ is an elementary extension of $\mathcal{M}$, can we immediately conclude that if $\mathcal{M} \models T$, then $\mathcal{N} \models T$? Why or why not?
My line of reasoning
We have for all $\mathcal{L}$-formula $\phi$, $\forall \bar{a} \in \mathbb{M}$ and its interpretation in $\mathcal{N}$ $\pmb{k}(\bar{a})$: $$\mathcal{M} \models \phi(\bar{a}) \Leftrightarrow \mathcal{N} \models\phi(\pmb{k}(\bar{a})).$$
Let $T$ be comprised of such $\phi$'s, then we have $\mathcal{N} \models T$ as desired.
Update
My question is a more general version of this question, which gives more context about what $T$ is.
Suppose $T$ has $\forall\exists$-axiomatization, $(\mathbb{I},<)$ is a linear order, and $(\mathcal{M}_i : i \in \mathbb{I})$ is a chain of models of $T$. We show that $\bigcup \mathcal{M}_i$ is a model of $T$.