Ultraproducts and types! Let $\mathcal{C}=\{A_k\}_{k<\omega}$ be a class of finite $\mathcal{L}$-structures. Let $p(x)=\{\varphi(x)\}_{i<\omega}$ be a consistent family of $\mathcal{L}$-formulas (you can think about $p(x)$ as a type), and let $\mathcal{M}=\prod_{i\in I}A_i/\mathscr{U}$ be an infinite ultraproduct of members of $\mathcal{C}$ (where $I$ is an infinite subset of  $\omega$, and $\mathscr{U}$ is a non-principal ultrafilter on $I$).  By the Łoś Theorem we know that for every $\mathcal{L}$-formula $\psi(x)$ and every $[(a_j)_{j\in I}]\in M$:
$$ \mathcal{M}\models \psi\big([(a_j)_{j\in I}]\big) \text{ iff }  \{i\in I: A_i\models \psi(a_i)\}\in \mathscr{U}.$$
Question (1):  What can we say if $\mathcal{M}\models p(\bar{a})$ for some $\bar{a}=[(a_j)_{j\in I}]\in M$?
Question (2): If for each $i<\omega $, $\mathscr{U}_i:=\{j\in \omega: A_j\models \varphi_i(a_j) \text{  for some } a_j\in A_j \}\in \mathscr{U}$, for some ultrafilter $\mathscr{U}$ on $\omega$. Can we conclude that there is a ultraproduct of members of $\mathcal{C}$ that realizes $p(x)$?
Question (3): In general, for a given type $p(x)=\{\varphi(x)\}_{i<\omega}$,  what are the sufficient conditions on members of $\mathcal{C}=\{A_k\}_{k<\omega}$ to conclude that "there is an ultraproduct $\mathcal{M}$ of members of $\mathcal{C}$ that realizes $p(x)$"?
(In general, it is not clear for me what happen when an ultraproduct satisfies a type. Any comment, idea or refrence would be appreciated)
 A: About (1): it means that for each $\varphi_i$, for $\mathscr U$-almost all $j$  you have $A_j\models \varphi_i(a_j)$. It does not at all imply that $p$ is realised by any tuple in any of the $A_j$, if that is what you were thinking. I don't think there is much more to be said, unless you have some more specific questions.
About (2): as stated, no, this is not true (indeed, the hypothesis, as you have put it, is always true, unless some $\varphi_i$ is not satisfied in any $A_k$ at all). You have probably meant to assume that there is a fixed $\mathscr U$ such that for all $i$, $\mathscr U_i\in \mathscr U$. This is no longer trivial, but it is still not true, not even if $\mathscr U_i=\omega$ for all $i$. To find a counterexample, consider the case when you are working in the abelian group language, and you have $\varphi_1(x)="x\neq 0"$, $\varphi_2(x)="x+x=0"$ and $\varphi_3(x)="x+x+x=0"$. If you assume that the analogue of this condition holds for finite conjunctions of $\varphi_i$-s, though --- then the answer is yes, by Łoś's theorem.
About (3), the most elegant criterion I can think of: such an ultraproduct exists if and only if the set $\{\varphi_n(x)\mid n\in \mathbf N\}$ is finitely satisfiable in $\{A_k\mid k\in \mathbf N \}$, in the sense that for any $N$, there is some $k$ such that $A_k\models \exists x \bigwedge_{n<N} \varphi_n(x)$.
