On the translation between a morphism of schemes and its geometric description Within the theory of schemes, rather than formally defining a morphism of schemes (saying which prime goes to which, and what the pullback map looks like), we often give the geometric intuition behind it. I am having difficulties translating between these two perspectives.
Easier example. Let's work in $\mathbb{R}$. I want to formalize what 'the projection of the parabola $y = x^2$ to the $y$-axis' means. Within the world of schemes, I guess that means you want a morphism from $\operatorname{Spec} \mathbb{R}[x,y] / (y - x^2)$ to $\operatorname{Spec} \mathbb{R}[y]$. Which one? It suffices to pick the right ring map $\mathbb{R}[y] \to \mathbb{R}[x,y] / (y-x^2)$. The obvious choice would be the inclusion map, so that's going to be what you want. To be honest the only reason I 'get' this is because this is the only reasonable choice.
Harder example. Consider the circle of radius $1$ around the origin. We want to find rational points on this circle. One of them is the point $p = (1,0)$. If $q$ is another rational point, consider the line going through $p$ and $q$, and compute the slope of this line. We end up with a map taking the coordinates $q = (x,y)$ and sending it to the slope $y/(x-1)$. Somehow this should yield a rational map from the circle to the affine line, i.e. a map $\operatorname{Spec}\mathbb{Q}[x,y] / (x^2 + y^2 - 1) \dashrightarrow \mathbb{A}_{\mathbb{Q}}^1$. Which rational map should it be? It beats me. Intuitively I understand what the map should do. I also understand that the formula we described gives the behaviour on closed points. But I'm not able to translate the intuition to a full fledged rational map.

Question. What should the map be in the harder example?

What I tried. The problem that I'm facing is that we are looking for a rational map, and not a morphism, so we cannot consider a corresponding map of rings. In this case, we're lucky that the birational 'inverse' of this map is an actual morphism, and this morphism should correspond to a ring map $\mathbb{Q}[x,y] / (x^2 + y^2 - 1) \to \mathbb{Q}[z]$. I can't think of a reasonable choice. One might also try and define the desired rational map on small affines and patch them together but this seems more tedious, and I end up getting stuck with the same problem.

Follow-up question. What is the general procedure for going from a geometric description to an actual morphism, and how do we know when
  it suffices to describe the behaviour on the closed points?

Edit. I would also be very happy with a reference. Unfortunately no book I know of seems to refer to this 'bridge'. Vakil formally defines morphisms of schemes but then proceeds to give 'examples' of morphisms of schemes by just providing coordinate maps.* Hartshorne and Mumford give examples in the world of varieties (where coordinate maps are fine because you only have the closed points to deal with) but forget about them when moving to schemes. Eisenbud gives only very basic ones where there's really only one reasonable choice of ring map. Liu does not really bother with examples.
*: That said, he is the only one who acknowledges the existence of the 'bridge'. I quote: "Suppose you are looking for rational points on the circle $C$ given by $x^2 + y^2 = 1$. One rational point is $p = (1,0)$. If $q$ is another rational point, then $pq$ is a line of rational (non-infinite) slope. This gives a rational map from the conic $C$ to $\mathbb{A}^1_{\mathbb{Q}}$, given by $(x,y) \mapsto y / (x-1)$." He then mentions: "Something subtle just happened: we were talking about $\mathbb{Q}$-points on a circle, and ended up with a rational map of schemes." But he doesn't say what this 'something subtle' is supposed to be, and that's precisely the detail that I've been desperately looking for for a long time. What is it? 
 A: 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.
A: 
Harder example. Consider the circle of radius $1$ around the origin. We want to find rational points on this circle. One of them is the point $p = (1,0)$. If $q$ is another rational point, consider the line going through $p$ and $q$, and compute the slope of this line. We end up with a map taking the coordinates $q = (x,y)$ and sending it to the slope $y/(x-1)$. Somehow this should yield a rational map from the circle to the affine line, i.e. a map $\operatorname{Spec}\mathbb{Q}[x,y] / (x^2 + y^2 - 1) \dashrightarrow \mathbb{A}_{\mathbb{Q}}^1$. Which rational map should it be? It beats me. Intuitively I understand what the map should do. I also understand that the formula we described gives the behaviour on closed points. But I'm not able to translate the intuition to a full fledged rational map.
Question. What should the map be in the harder example?

It looks like you are getting confused by the points at infinity. There is an isomorphism of any smooth quadratic curve in $\mathbb{P}^2$ to $\mathbb{P}^1$ but you are taking the circle, which has two complex points at infinity and comparing it to the line with one real point at infinity: these affine spaces are not isomorphic. Thus$$t={y\over{{x-1}}}$$is indeed an isomorphism of the field $k(t)$ and the quotient field of $$k[x,y]/(x^2+y^2-1)$$---here is your homework: to check this, find the expressions for $x$ and $y$ as rational functions of $t$. But the point $(1,0)$ on the circle goes to $t=\infty$.
