A graph in descriptive complexity - is $x$ already a vertex?
(First-order logic) formulae are independent of structures; you just have variables, predicates and possibly functions (in addition to the syntactical symbols). To decide whether a formula $\varphi$ is true in a given structure $S$ (consisting of a universe of individuals $U$ and an interpretation of predicates $I$), i.e. whether $S$ is a model of $\varphi$, you apply the semantic rules of first order logic.

Concerning your specific question: if you have $\varphi = \forall x: F$ where $F$ is some formula, then $x$ denotes exactly all vertices iff you defined your structure $S = (U,I)$ such that $U$ contains exactly all vertices. Note that if you do that and want access to the edges, then you'll need to define some predicate $e(x,y)$ such that $e(x,y) \Leftrightarrow \text{there is an edge between } x \text{ and } y$.

More generally: the quantifiers do not differentiate between elements of your universe, the predicates will have to do that. If you want to check $F$ only for all vertices but your universe contains elements that are not vertices, then you do what you suggested: introduce a predicate $v(x)$ and use $\varphi' = \forall x: (v(x) \rightarrow F)$