If you are looking for good first examples, Mumford's Red Book and Eisenbud and Harris's 'Geometry of Schemes' have some good pictures and examples.
Its worth playing around with Spec(O_c), where O_c is the ring of integers in the extension of Q by the square root of c, and thinking about it as a scheme over Spec(Z). In particular, several somewhat mysterious number theory terms like 'ramified' and 'split' make geometric sense in this context.
Its also worth thinking about what the p-adics should look like as a scheme - a formal neighborhood of p in Spec(Z) (though to make this precise you need to know what formal schemes are).
Its also not a terrible idea to pick up a book on algebraic number theory and try to translate everything that is said into a geometric statement (the trick is to realize every time talk about a field, they are really talking about the ring of integers in that field).
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).)
Best Answer
First, the Fontaine-Winterberger isomorphism can also be recovered from a theorem of Deligne, namely Thm 2.8 here. Deligne showed that if two local fields $K_1$ and $K_2$ (possibly of different characteristic) were such that $O_{K_1}/\mathfrak{m}_{K_1}^N \simeq O_{K_2}/\mathfrak{m}_{K_2}^N$ then the category of finite etale extensions of $K_1$ with ramification bounded by $N$ is isomorphic to the corresponding category for $K_2$. One gets the Fontaine-Winterberger isomorphism by taking $K_1 = \mathbb{Q}_p(p^{\frac{1}{p^N}})$, $K_2 = \mathbb{F}_p((t^{\frac{1}{p^N}}))$ and letting $N \rightarrow +\infty$.
Abbes and Saito gave a geometric description of the ramification filtration of local fields in this paper. They were motivated by the case of imperfect residue fields, but a quick survey of their idea in the classical case can be found in Section $2$ of this article by Shin Hattori.
Shin Hattori used Abbes-Saito's idea in order to reprove and extend Deligne's result. This yields a geometric proof of Deligne's theorem, using rigid geometry and perfectoid spaces. The basic idea is that the ramification of $K_1$, resp. $K_2$, can be read from the connected components of some rigid spaces $X_1^{\leq j}$, $X_2^{\leq j}$ defined by Abbes-Saito, and if $j \leq N$ one can exhibit explicit pro-etale covers $\widetilde{X}_1^{\leq j}$, $\widetilde{X}_2^{\leq j}$ such that $\widetilde{X}_1^{\leq j}$ is the tilt of $\widetilde{X}_2^{\leq j}$. More details can be found in this survey by Shin Hattori, or in the paper here.
Remark: the only thing needed about perfectoid spaces is the fact that tilting preserves connected components. This is even easier than showing that the corresponding adic spaces are homeomorphic, and in any case does not use the equivalence of etale site (so no circular reasoning here). For the sake of completeness: if $A$ is a ring then idempotents of $A$ are in bijection with idempotents of the monoid $A^{\flat} = \lim_{x \mapsto x^p} A$.