Geometry of the subalgebra $\Bbbk[x^2-1]\leq \Bbbk[x]$ (intuition for integral elements) Given a field $\Bbbk$ consider the subalgebra $\Bbbk[x^2-1]\leq \Bbbk[x]$. This is an integral extension of algebras. Write $\mathfrak q= (x-1)\vartriangleleft \Bbbk[x]$ and $\mathfrak p=\Bbbk [x^2-1]\cap \mathfrak q\vartriangleleft \Bbbk[x^2-1]$. The extension $\Bbbk[x^2-1]_{\mathfrak p}\hookrightarrow \Bbbk[x]_\mathfrak{q}$ is not integral, because $\frac 1{x+1}\in \Bbbk[x]_\mathfrak{q}$ is not integral.
I would like to understand the geometry behind this example. I know being integral is local on the target, stable under base change, and has the fundamental consequences of being universally closed and surjective with finite fibers. I also know being integral behaves well w.r.t closed subschemes (quotients by ideals). In fact since the algebras in question are of finite type, 'integral' is equivalent to 'finite', so I don't mind pretending to deal with surjective finite ramified covering maps (though it would be nice if this were unnecessary for this problem).
So geometrically, we have the surjective ramified covering map $\mathbb A^1_\Bbbk\twoheadrightarrow \mathbb A^1_\Bbbk$ with two branches. Somehow the picture is not as nice when we zoom into the smallest open neighborhood of $\mathfrak q\in \mathbb A^1_\Bbbk$ upstairs and the smallest open neighborhood of its image $\mathfrak p\in \mathbb A^1_\Bbbk$ downstairs. Now intuition misleads me: the reason that restriction of covering maps to open subsets of the domain are not generally covering maps is that you can "cut off" fibers upstairs. On the other hand, I'd think that localizing to a small enough open neighborhood of $\mathfrak q\in \mathbb A^1_\Bbbk$ would in particular localize to just one branch, and then the map would basically be an isomorphism. Of course this cannot be true since I have proved $\Bbbk[x^2-1]_{\mathfrak p}\hookrightarrow \Bbbk[x]_\mathfrak{q}$ is not even integral, not to mention an isomorphism.
Question 1. What's the problem with my intuition above?
Trying to think more concretely, taking $\Bbbk=\mathbb R$ is misleading since $x\mapsto x^2-1$ does not "look" surjective, so I tried to look at $\Bbbk=\mathbb C$. I think I  understand of the topology of complex function $z\mapsto z^2-1$ map, but I certainly don't know how to think of the stalks/smallest neighborhoods topologically.
Question 2. Which topological property of "ramified covering maps" fails after localization?
Finally:
Question 3. What is the geometric meaning of the particular element $\frac 1{x+1}$ not being integral?
 A: Well, you seem to have forgotten how the Zariski topology works.  Imagine we have the branched covering map $z\mapsto z^2$ on $\mathbb{C}$ (I find it easier to think about if you translate so it's $z^2$ instead of $z^2-1$; after all, $\mathbb{k}[x^2-1]=\mathbb{k}[x^2]$).  This should be a very familiar picture from complex analysis: you have the complex plane, and you're doubling the argument at every point so that it winds around the origin twice as fast.
Now, you want to restrict this branched cover to open neighborhoods of the point $1$ in the domain, but we are working with the Zariski topology, not the Euclidean topology!  So, an open neighborhood is just a cofinite subset of $\mathbb{C}$.  Obviously if we remove finitely many points, we're not going to get just a single branch of our cover.  No matter how many remove, we will still have points that are arbitrarily close to the other point $-1$ which has the same image as our point $1$.  The localizations are a direct limit of considering these Zariski open sets, so it shouldn't be surprising that they are no better.
In particular, what's going on with $\frac{1}{x+1}$?  Well, it's exactly what you guessed: the restriction of a covering map on the domain may no longer be a covering map, because we cut off a fiber.  In particular, we are allowed to remove $-1$ from the fiber upstairs when considering neighborhoods of $1$, but we still have its image downstairs, so there is a point "missing" upstairs.  And we can't remove the entire branch of that missing point, since in the Zariski topology we can only remove finitely many points at a time.
