On the translation between a morphism of schemes and its geometric description

For clarity let me write $\mathbb Q[X,Y]$ for the ring of polynomials and $\mathbb Q[x,y]=\mathbb Q[X,Y]/\langle X^2+Y^2-1\rangle$ for the ring of regular functions on your circle $S:=V(X^2+Y^2-1)\subset \mathbb A^2_\mathbb Q=\operatorname {Spec}(\mathbb Q[X,Y])$

Your rational map can be described in two different ways:
a) It corresponds to the isomorphism of the rational function fields of the two varieties $\mathbb A^1_\mathbb Q$ and $S$: $$\mathbb Q(z)=\operatorname {Rat}(\mathbb A^1_\mathbb Q)\stackrel {\cong}{\to} \operatorname {Frac}(\mathbb Q[x,y]) = \operatorname {Rat}(S):z\mapsto \frac{y}{x-1}$$

b) It corresponds to the regular map $$f:U\to \mathbb A^1_\mathbb Q$$ defined only on the open subset $U=S\setminus \{(1,0)\}=\operatorname {Spec}(\mathbb Q[x,y])\setminus V(x-1)\subset S$
(Recall that a rational map $X\dashrightarrow Y$ between varieties is given by a regular map $U\to Y$defined on a non-empty open subset $U\subset X)$
This morphism $f$ is induced by the ring morphism $$f^*:\mathbb Q[z] \to\mathcal O(U)=\mathbb Q[x,y]_{( x-1)}=\mathbb Q [x,y,\frac {1}{x-1}]:z\mapsto \frac{y}{x-1} $$

Remark
Your puzzlement seems to be due to the misconception that the morphism of schemes $f$ would correspond to a ring morphism $f^*:\mathbb Q[z]\to \mathbb Q[x,y]$: this is impossible because the codomain would be too small.