The following is the definition of disjoint union spaces in John Lee's "Introduction to Topological Manifolds":
Let $(X_{\alpha})_{\alpha\in A}$ be an indexed family of nonempty topological spaces. We define the disjoint union topology on $\amalg_{\alpha\in A}X_{\alpha}$ by declaring a subset of the disjoint union to be open if and only if its intersection with each set $X_{\alpha}$ (considered as a subset of the disjoint union) is open in $X_{\alpha}$. With this topology, $\amalg_{\alpha\in A}X_{\alpha}$ is called a disjoint union space.
In this definition, what does it mean for a set to be open in $X_{\alpha}$? Here the disjoint union $\amalg_{\alpha\in A}X_{\alpha}$ is defined by $$\amalg_{\alpha\in A}X_{\alpha}=\{(x,\alpha):\alpha\in A, x\in X_{\alpha}\},$$
and by "considered as a subset of the disjoint union", he means that he is identifying the set $X_{\alpha}$ with the set $$\{(x,\alpha): x\in X_{\alpha}\}.$$
Best Answer
For every index $\alpha \in A$ there is a canonical inclusion map $\sigma_\alpha :X_\alpha \to \coprod_{\alpha \in A} X_\alpha$ given by $x \mapsto (x,\alpha)$. A subset $V$ of $\coprod_{\alpha \in A} X_\alpha$ is declared as open if $\sigma_\alpha^{-1}(V)$ is open in $X_\alpha$ for every $\alpha \in A$. More precisely, we define a topology $\mathfrak{T}$ on $\coprod_{\alpha \in A} X_\alpha$ as follows: $$ \mathfrak{T} = \left\{ V \subseteq \coprod_{\alpha \in A} X_\alpha : \sigma_\alpha^{-1}(V) \text{ is open in $X_\alpha$, for all $\alpha$} \right\}. $$
As you pointed out, we then identify $X_\alpha$ with the set $$ Y_\alpha := \{ (x,\alpha) : x \in X_\alpha\}. $$ Then, $\sigma_\alpha^{-1}(V)$ can be viewed as the "intersection of $V$ with $X_\alpha$". Indeed, after identifying $X_\alpha$ with $Y_\alpha$ above, the pre-image $\sigma_\alpha^{-1}(V)$ asks which points in $V$ live in $X_\alpha$.