Can I glue the X axis to the Y axis? Consider the following diagram of algebraic varieties:
$$\mathbb{A}^0 \to \mathbb{A}^1 \rightrightarrows \mathbb{A}^2$$
Here the first arrow is the inclusion of the origin into the line, and the next two are the inclusion of the line into the plane as the X and the Y axes.

Does this diagram have a colimit in the category of schemes?

(The first arrow giving inclusion of the point is not relevant; it just so happens that it was there when I met this question.)
 A: Dan Petersen has already answered the hard part -- how to show that the affine pushout $\{ f \in k[x,y] : f(t,0) = f(0,t) \}$ is also the scheme push out. I write to record a discussion in the comments about how to see that this ring is finitely generated and obtain an explicit list of generators.
Let $R = k[x,y]$ and let $S = k[x,y]^{S_2} = k[b,c]$ where $b=x+y$ and $c = xy$. Let $A = \{ f \in k[x,y] : f(t,0) = f(0,t) \}$. Then clearly $A$ is an $S$-submodule of $R$. Since $S$ is noetherian and $R$ is generated as an $S$-module by $1$ and $x$, this shows that $A$ is finitely generated as an $S$-module.
If the characteristic of $k$ is not $2$, one can get explicit generators easily. $A$ is invariant under switching the generators $x$ and $y$, so $A$ splits into positive and negative eigenspaces, call them $A_+$ and $A_-$, for this switch. The positive eigenspace is just $S$. The negative eigenspace $A_-$ is $xy(x-y) S$, since it is easy to see that anything in $A_-$ is divisibly by $xy(x-y)$, and the quotient is in $A_+$. So $A = S \oplus xy(x-y) S$ and the ring is generated by $b = x+y$, $c = xy$ and $f=xy(x-y)$, with the defining relation $f^2 = c^2 (b^2-4c)$.
A: We can rewrite the coequalizer as the pushout of the diagram
$$
\begin{array}{ccc}
    X & \to & \mathbb A^2 \\
    \downarrow & & \\
    \mathbb A^1 & &  \\
\end{array}
$$
where $X$ is the union of the $x$- and $y$-axis, and the vertical map quotients by the involution swapping the two components.
The category of affine schemes has all pushouts: they are given by fibered product of coordinate rings. But a pushout in the category of affine schemes is not necessarily a pushout in the category of all schemes. A sufficient condition for a pushout of affine schemes to be a pushout in the category of schemes can be found in a paper of Karl Schwede ("Gluing schemes and a scheme without closed points", Theorem 3.4): it suffices that one leg of the pushout is a closed immersion. So we are fine.
As noted by Gro-Tsen in a comment, the pushout can be written explicitly as $\mathrm{Spec}(R)$ where $R = \{ f \in k[x,y] : f(t,0)=f(0,t)\}$.
