I use these definition, but I need set-theoretic definitions because I define $h: A\to B \text{ for }\left\{\begin{array}{l} h \subseteq (A \times B) \\ \forall x,y,z:(((x,y) \in h \wedge (x,z) \in h) \to y=z) \\ \operatorname{dom}(h)=A \\ \operatorname{cod}(h)\subseteq B\end{array} \right.$
For example: $f_1 + f_2:=\{x|\exists y,z : (x=(y,z) \wedge y \in V \wedge z= f_1 + f_2)\}$ Is correct? If yes, I demonstrate $f_1 + f_2: V \to W$...
If $f: V \to W$ and $g: V \to W$ are maps from a set $V$ to a set $W$ and $\bot$ is a law of $W$, we ca define $f \bot g$ by :
$$f \bot g: V \to W, x \mapsto f(x) \bot g(x)$$
If you want use the theoritic definition you can see $f$ as a part of $V \times W$ such that :$$(\star\star) \quad\left\{\begin{array}{l}(\forall x \in V)(\exists y \in W) \quad (x,y)\in f \\ (\forall(x,y,y') \in V \times W \times W)\quad (x,y) \in f \wedge (x,y') \in f \Rightarrow y=y' \end{array} \right.$$
Then the définition of $f+g$ is :
$$f+g=\{(x,y)/ (x \in V) \wedge (y \in W) \wedge (y=y_1\bot y_2) \wedge ((x,y_1)\in f) \wedge ((x,y_2) \in g)\}$$
Well, I don't think you're being honest when you say you see a function like a subset of the cartesian product. Because the example you gave is not a subset of any cartesian product. What I think is that you have a homework problem to give a set-theoretic representation of the sum of two functions.
First, it's important to understand what it means to represent a function as a subject of the cartesian product of domain and target. If $f\colon X \to Y$ is a function, we can create the set $$ R = \left\{(x,y) \mid f(x) = y \right\} $$ Since $R\subset X\times Y$, this is a relation. If $X$ and $Y$ are subsets of $\mathbb{R}$ and you plot $f$ in the plane, you see that $R$ is just the graph of $f$.
The defining property of a function is that its graph passes the “vertical line test”—that is, any vertical line intersects the graph in at most one point. Let's translate that to a set-theoretic condition. If the vertical line $x=x_0$ intersected the graph of $f$ in two points, then there would be two distinct points $(x_0,y_1)$ and $(x_0,y_2)$ on the graph of $f$. That is, the points $(x_0,y_1)$ and $(x_0,y_2)$ would both be in $R$. So $R\subset X\times Y$ is a function if there is no points $x_0$ and $y_1 \neq y_2$ with $(x_0,y_1)$ and $(x_0,y_2)$ in $R$. Contrapositively, $R$ is a function if, for any points $x_0$, $y_1$, and $y_2$ with $(x_0,y_1)$ and $(x_0,y_2)$ in $R$, then $y_1 = y_2$.
Now to your problem. You have $X$ and $Y$ vector spaces and two linear maps $f_1$ and $f_2$ from $X$ to $Y$. Take a pair $(x_0,y_0)$ in $X\times Y$. Under what conditions is $y_0=(f_1+f_2)(x_0)$? It would have to be true that $y_0 = y_1 + y_2$ where $y_1 = f_1(x_0)$ and $y_2 = f_2(x_0)$. In other words, $y_0 = y_1 + y_2$, where $(x_0,y_1)$ is in $R_{f_1}$, and $(x_0,y_2)$ is in $R_{f_2}$.
