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 $f^2 = c^2 (b^2-4c)$.

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