A subset $S \subset \mathbb{R}^3$ is a regular surface if for each $p \in S$ there exists a neighborhood $V \subset \mathbb{R}^3$ and a map $$\mathbf{x} : U \rightarrow V \cap S$$ where $U$ is an open set in $\mathbb{R}^2$ and the map satisfies certain properties. My question is, is this really that different from saying that there is a map that takes a point in $U$ and maps it to a point $p \in S$?
I am not really understanding the whole $V \cap S$ part. If $V$ is not in $S$ then the definition makes no sense. Why is there a need for this set $V$? Why can't we just say $$\mathbf{x} : U \rightarrow S?$$
As a subset of $\Bbb R^3, S$ can be given the subspace topology, and the existence of homoeomorphism $ \phi : U \subseteq \Bbb R^2 \to V \cap S$ (which is an open set) means that $S$ is actually a manifold, and if it is a diffeomorphism then a surface. This “certain conditions” are very important in that it helps you talk about continuous and smooth maps, properties such compactness, connectedness, and the notion of tangent spaces, curvatures etc. None of this can be done with a map that just maps a point in $U$ to a point in $S$.
Even though $V$ is not in $S$, $V \cap S$ is and hence the definition makes sense. If you keep in mind that $S$ has a topology and hence open sets, $ \phi $ can be thought of as map into $W \subseteq S, W$ open in $S$.