I am wondering if there is any algebraic definition for the graph of a vector field.
I know that for functions $f:\mathbb{R}^2\rightarrow \mathbb{R}$, the graph of $f$ is given by $\{(x,y,z)\in\mathbb{R}^3\mid z=f(x,y)\}$. So, for example, the graph of the function $f(x,y)=x^2+y^2$ is given by $\{(x,y,z)\in\mathbb{R}^3\mid z=x^2+y^2\}$.
My question is how to describe the graph of a vector field. Say we have the function $f:\mathbb{R}^2\rightarrow \mathbb{R}^2$ given by $f(x,y)=(x^2+x-2y^2+1,-x^2+y^2+3y-2)$. I believe the graph of this function should sit inside $\mathbb{R}^2\times\mathbb{R}^2$, however, I am confused how to express this similarly to the example given above. I am ultimately trying to find the intersection of the graph of this function with the set of points in $\mathbb{R}^2\times \mathbb{R}^2$ of the form $\{(x,y,x,y)\}$. Is there a way to do this just with the expression given above?
Given the function $f : \mathbb{R}^2 \to \mathbb{R}$, the graph of $f$ is given by $\{(x,y,z)\in\mathbb{R}^3\mid z=f(x,y)\}$. This is because we identify the image of $f$ with a third variable, namely $z$, sitting in another copy of $\mathbb{R}$.
Now consider $f:\mathbb{R}^2\rightarrow \mathbb{R}^2$ given by $f(x,y)=(x^2+x-2y^2+1,-x^2+y^2+3y-2)$, as in your question. This has two values in its image, so you are correct that the graph sits in $\mathbb{R}^{2} \times \mathbb{R}^{2} = \mathbb{R}^{4}$. If we write $f(x, y) = (z, w) \in \mathbb{R}^{2}$, then the graph of $f$ is the set $$ \{(x, y, z, w) \in \mathbb{R}^{4} \mid z = x^2+x-2y^2+1, w = -x^2+y^2+3y-2\}. $$ If you want to intersect this with all points $(x, y, x, y)$, then you want $(x, y, z, w) = (x, y, x, y)$ so you are solving $z = x$ and $w = y$. That is, you are solving $$ x = x^2+x-2y^2+1, \hspace{20pt} y = -x^2+y^2+3y-2. $$