The scheme $X_1$ is nothing but the affine line: $X_1 = Spec(A[x]) = \mathbb{A}^1$.
The scheme $X_2$ is nothing but two copies of the affine line: $X_2 = Spec(A[y]\oplus A[z]) = \mathbb{A}^1\sqcup\mathbb{A}^1$.
Your morphism $X_2\rightarrow X_1$ corresponds to the morphism $A[y]\oplus A[z]\rightarrow A[x]$ given by sending both $y$ and $z$ to $x^2$. In geometric terms, on both pieces of $\mathbb{A}^1$ in $X_2$, it's given by $x\mapsto x^2$.
There is no similar morphism $X_1\rightarrow X_{1/2}$, because the scheme $X_1$, unlike $X_2$, is irreducible.
The only thing you can do is to give another morphism $\mathbb{A}^1\mapsto\mathbb{A}^1$ sending $x$ to $x^2$.
In your language, this means the following: let $X_{1/2}$ be the same as $X_1$, i.e. $Spec(A[x, y]/(y-x^2))$. Then there is a morphism $X_1\rightarrow X_{1/2}$ given (in terms of points) by $(x, y)\mapsto(x^2, y^2)$.