Let $x$ be a nonempty set and $w' \in x$. Then by the axiom schema of specification:
\begin{equation} \exists u\forall v(v \in u \Longleftrightarrow v \in w' \land \forall w(w \in x \implies v \in w)). \end{equation}
Am I correct in thinking of this set as $\bigcap x$?
I think that the answer is yes, but your formalization is lacking.
You seem to want to define $\bigcap x$, so you are looking for a formula $\varphi(u,x)$ such that $\varphi(y,x)$ holds if and only if $y\in\bigcap x$.
Indeed, this formula should state that $y$ is a member of all the elements of $x$: $$\varphi(y,x):=\forall w(w\in x\rightarrow y\in w)$$
Now to prove that this is indeed the intersection and that it is a set we appeal to the fact that $x$ is not empty. Take any $w'\in x$, then $\{y\in w'\mid\varphi(y,x)\}$ is a set by specification and it equals to $\bigcap x$.
Next you should prove this does not depend on the choice of $w'$, and therefore it is meaningful to denote this as $\bigcap x$, as it is a function of $x$ independent of the choice of $w'$.