The intuition for this method of passing from a rational solution to an integral solution seems pretty simple to me: passing from a rational solution to a nearby integral point (not necessarily a solution) is passing to a point whose denominators are 1, so you can anticipate that when you intersect the line through your rational solution and the nearby integral point with whatever curve or surface contains your solutions, the second intersection point on that line will have denominators that have moved closer to 1. That is, connecting a rational solution with some integral point will spit out a new solution whose denominators are somewhere between the denominators of your solution and the denominators of the integral point you used to produce the line.
Of course intuition is one thing and checking the details is another: you choose the integral point nearby and the math has to work out to show the denominators really get smaller in the second solution you produce. For instance, this method of proving the 3-square theorem goes through without a problem for a similar 2-square theorem (if an integer is a sum of two rational squares than it's a sum of two integral squares by the same method, replacing the sphere x^2 + y^2 + z^2 = a with the circle x^2 + y^2 = a). But this intuitive way of creating an integral solution from a rational solution breaks down if you apply it to the 4-square theorem: the inequalities in the proof just barely fail to work (sort of like doing division with remainder and finding the remainder is as big as the divisor instead of smaller).
The intuition also breaks down if you slightly change the expression x^2 + y^2 (sticking to two variables). Consider x^2 + 82y^2 = 2 and the rational solution (4/7,1/7). Its nearest integral point in the plane is (1,0), and the line through these intersects the ellipse in (16/13,-1/13), so the denominator has gone up. There actually are no integral solutions to x^2 + 82y^2 = 2. Or if we take x^3 + y^3 = 13 and the rational solution (2/3,7/3), its nearest integral point in the plane is (1,2), the line through these meets the curve again in (7/3,2/3), whose nearest integral point in the plane is (2,1), the line through them meets the curve in (2/3,7/3),...
A few years ago when I was giving some lectures on the method of descent, I worked out some examples of this geometric "three-square" theorem (start with an equation a = x^2 + y^2 + z^2 where a is an integer and x, y, and z are rational and produce in a few steps an equation where x, y, and z are integral) and I noticed in my initial examples that the denominators in each new step did not merely drop, but dropped as factors, e.g., if the common denominator at first was 15 then at the next step it was 5 and then 1. Maybe the denominators always decreas through factors like this? Nope, eventually I found a case where they don't: if you start with
13 = (18/11)^2 + (15/11)^2 + (32/11)^2
then the integral point nearest (18/11,15/11,32/11) is (2,1,3) and the line through these two points meets the sphere 13 = x^2 + y^2 + z^2 in the new point (2/3,7/3,8/3), so the denominator has fallen from 11 to 3, which is not a factor. (At the next step you will terminate in the integral solution (0,3,2).)
Revised
Interesting question.
Here's a thought:
You can think of a ring, such as $\mathbb Z$, in terms of its monoid of affine endomorphisms
$x \rightarrow a x + b$. The action of this monoid, together with a choice for 0 and 1, give the structure of the ring. However, the monoid is not finitely generated, since the
multiplicative monoid of $\mathbb Z$ is the free abelian monoid on the the primes, times
the order 2 group generated by $-1$.
If you take a submonoid that uses only
one prime, it is quasi-isometric to a quotient of the hyperbolic plane by an action of $\mathbb Z$, which is multiplication by $p$ in the upper half-space model. To see this, place a dots labeled by integer $n$ at position $(n*p^k, p^k)$ in the upper half plane, for every pair of integers $(n,k)$, and connect them by horizontal line segments and by vertical line segments whenever points are in vertical alignment. The quotient of upper half plane by the hyperbolic isometry $(x,y) \rightarrow (p*x, p*y)$ has a copy of the Cayley graph for this monoid. This is also quasi-isometric to the 1-point union of two copies of the hyperbolic plane, one for negative integers, one for positive integes. It's a fun exercise,
using say $p = 2$. Start from 0, and recursively build the graph by connecting $n$ to $n+1$ by one color arrow, and $n$ to $2*n$ by another color arrow. If you arrange positive integers in a spiral, you can make a neat drawing of this graph (or the corresponding graph for a different prime.) The negative integers look just the same, but with the successor arrow reversed.
If you use several primes, the picture gets more complicated. In any case, one can take rescaled limits of these graphs, based at sequence of points, and get asymptotic cones for the monoid. The graph is not homogeneous, so there is not just one limit.
Another point of view is to take limits of $\mathbb Z$ without rescaling, but
with a $k$-tuple of constants $(n_1, \dots , n_k)$. The set of possible identities among polynomials in $k$ variables is compact, so there is a compact space of limit rings for $\mathbb Z$ with $k$ constants. Perhaps this is begging the question: the identitites that define the limits correspond to diophantine equations that have infinitely many solutions.
Rescaling may eliminate some of this complexity.
A homomorphism $\mathbb Z[x,y,\dots,z]/P$ to
$\mathbb Z$ gives a homomomorphism of the corresponding monoids, so an infinite sequence of these gives an action on some asymptotic cone for the affine monoid for $\mathbb Z$.
With the infinite set of primes, there are other plausible choices for how to define length; what's the best choice depends on whether and how one can prove anything of interest.
Best Answer
This paper by Christian Elsholtz seems to be exactly what you're looking for. It motivates the Zagier/Liouville/Heath-Brown proof and uses the method to prove some other similar statements. Here is a German version, with slightly different content.
Essentially, Elsholtz takes the idea of using a group action and examining orbits as given (and why not -- it's relatively common) and writes down the axioms such a group action would have to fulfill to be useful in a proof of the two-squares theorem. He then algorithmically determines that there is a unique group action satisfying his axioms -- that is, the one in the Zagier proof. The important thing is that having written down these (fairly natural) axioms, there's no cleverness required; finding the involution in Zagier's proof boils down to solving a system of equations.