First of all, the modal logic we are working with in this case is the basic one: that is, all propositional formulas, plus formulas of the form $\Diamond\phi$, where $\phi$ is any modal formula (we define $\Box\phi$ as $\neg\Diamond\neg\phi$).
Let's define satisfaction. For that, we must define models for modal logic. Given $M=(F,V),F=(W,R)$ and a set, $\Phi$, we say $M$ is a model for the modal logic based on $\Phi$ if $W$ is a set, $R\subseteq W\times W$, and $V$ is a function from $\Phi$ to $2^{W}$, that is, it maps each element of $\Phi$ to a subset of $W$. From this, we also define (among other things) that $\Phi$ is the set of propositional letters of the logic, $F$ is a frame for the same modal logic and $V$ is a valuation on that frame.
Now we can proceed with the satisfaction definition. Given $w\in W$, we say $M,w\Vdash \phi$, that is, $M$ satisfies $\phi$ in $w$, if the following is true: $\phi=p$, where $p\in \Phi$ and $w\in V(p)$, or $\phi=\neg \psi$ and $M,w\nVdash \psi$, or $\phi=\psi\land\chi$ and $M,w\Vdash\psi$ and $M,w\Vdash\chi$, or $\phi=\Diamond \psi$ and there exists $v\in W$ such that $Rwv$ and $M,v\Vdash\psi$.
Then, we can define validity. We say that $F\Vdash\phi$, that is, $\phi$ is valid in $F=(W,R)$, if, for all $w\in W$ and all valuations $V$ on $F$, $(F,V),w$ satisfies $\phi$.
A modal formula $\phi$ defines a class of frames $\textrm{F}$ if, for every frame $F$, $F$ is in $\textrm{F}$ if and only if $F\Vdash\phi$, that is, $\phi$ is valid in $F$. A similar definition applies to first-order logic, obviously replacing $\Vdash$ with $\vDash$. We say that a modal formula $\phi$ defines a first-order condition $\psi$ if they define exactly the same class of frames.
As part of this question, we were already given a hint on how to proceed; it is known that, in the frame $(\{u\}\cup u \cup \mathbb{N}, \ni)$, where $u$ is a non-principal ultrafilter on $\mathbb{N}$, $\Diamond\Box p \rightarrow \Diamond (\Box (p\land q) \lor \Box(p\land\neg q))$ is valid, and the original question suggests that you should establish that in any countable structure elementarily equivalent to the given frame the formula is not in fact valid.
The proof of the first fact (formula is valid on frame) does give out some intuition on how to proceed: The crucial step of said proof relies on any subset of $\mathbb{N}$ (or its complement) being a member of $u$ and that $u$ is closed under intersection, but we can expect a countable structure to be missing several of these elements, allowing the formula to be falsified.
Unfortunately, i have not been able to convert said intuition into an actual proof. Is there a way to do so? Or, alternatively, is there a different approach to this question?