What is the intuition behind uniform continuity? There’s another post asking for the motivation behind uniform continuity. I’m not a huge fan of it since the top-rated comment spoke about local and global interactions of information, and frankly I just did not get it. 

Playing with the definition, I want to say uniform continuity implies there’s a maximum “average rate of change”. Not literally a derivative, but the rate of change between two points is bounded in the domain. I’m aware that this is essentially Lipschitz continuity, and that Lipschitz implies uniform. This implies there’s more to uniform continuity than just having a bounded average rate of change. 

And also, how is it that $ f(x)=x$ is uniform yet $f(x)f(x)=g(x)=x^2$  is not? I understand why it isn’t, I can prove it. But I just don’t understand the motivation and importance of uniform continuity.
The real "gist" of continuity, in its various forms, is that it's the "property that makes calculators and measurements useful". Calculators and measurements are fundamentally approximate devices which contain limited amounts of precision. Special functions, like those which are put on the buttons of a calculator, then, if they are to be useful, should have with them some kind of "promise" that, if we only know the input to a limited amount of precision, then we will at least know the output to some useful level of precision as well.

Simple continuity is the weakest form of this. It tells us that if we want to know the value of a target function $f$ to within some tolerance $\epsilon$ at a target value $x$, but using an approximating value $x'$ with limited precision instead of the true value $x$ to which we may not have access or otherwise know to unlimited precision, i.e. we want

$$|f(x) - f(x')| &lt; \epsilon$$

then we will be able to have that if we can make our measurement of $x$ suitably accurate, i.e. we can make that

$$|x - x'| &lt; \delta$$

for some $\delta &gt; 0$ which may or may not be the same for every $\epsilon$ and $x$.

Uniform continuity is stronger. It tells us that not only do we have the above property, but in fact the same $\delta$ threshold on $x'$'s accuracy will be sufficient to get $\epsilon$ worth of accuracy in the approximation of $f$ no matter what $x$ is. Basically, if the special function I care about is uniform continuous, and I want 0.001 accuracy, and the max $\delta$ required for that is, say, 0.0001, by measuring to that same tolerance I am assured to always get 0.001 accuracy in the output no matter what $x$ I am measuring. If, on the other hand, it were the case that the function is merely continuous but not uniformly so, I could perhaps measure at one value of $x$ with 0.0001 accuracy and that accuracy would be sufficient to get 0.001 accuracy in the function output, but if I am measuring at another, such a tolerance might give me only 0.5 accuracy - terrible!

Lipschitz continuity is even better: it tells us that the max error in approximating $f$ is proportional to that in approximating $x$, i.e. $\epsilon \propto \delta$, so that if we make our measurement 10 times more accurate, say (i.e. one more significant figure), we are assured 10 times more accuracy in the function (i.e. gaining a significant figure in the measurement lets us gain one in the function result as well).

And in fact, all the functions (that are real-analytic, not combinatorial functions like nCr and what not) on your real-life calculator are at least locally Lipschitz continuous, so that while this proportionality factor (effectively, absolutely how many sig figs you get for a given number of such in the input) may not be the same everywhere, you can still be assured that in relative terms, adding 10x the precision to your measurements, i.e. one more significant figure, will always make the approximation (however good or not it actually is) returned by your calculator 10x more accurate, i.e. also to one more significant figure.

And to top it all off, all these forms of continuity - at least in their local variants, that is, over any bounded interval - are implied by differentiability.
One interpretation that I love is the one using non-standard analysis:

Let $f : E \subseteq \mathbb{R} \to \mathbb{R}$ be a function. Then the followings are equivalent:

$f$ is uniformly continuous.
Its hyperreal version ${{}^*\!f} : {{}^*\!E} \subseteq {}^*\mathbb{R} \to {}^*\mathbb{R}$ is continuous, i.e., whenever $x, y \in {{}^*\!E}$ are infinitely close, ${{}^*\!f}(x)$ and ${{}^*\!f}(y)$ are infinitely close as well.


Here, the 'hyperreal version' refers to the $*$-transform of $f$. The catch is that the domain ${{}^*\!E}$ of ${{}^*\!f}$ contains hyperreal numbers which are either infinitely close to $E$ or infinitely large. So the continuity needs to be tested around those numbers as well to establish uniform continuity.
This statement can be recast in terms of real numbers only:

Let $f : E \subseteq \mathbb{R} \to \mathbb{R}$ be a function. Then the followings are equivalent:

$f$ is uniformly continuous.

For any sequences $(a_n)$ and $(b_n)$ in $E$ such that $|a_n - b_n| \to 0$, we have $|f(a_n) - f(b_n)| \to 0$.



In this version, the role of infinitely close hyperreal numbers is replaced by a pair of sequences which become arbitrarily close. Again, we see that uniformly continuity is really about enforcing continuity infinitesimally beyond $E$. Here are some examples that demonstrate this idea:
Example 1. Let $f(x) = 1/x$ on $(0, \infty)$. If we pick two 'infinitesimals' $a_n = \frac{1}{n}$ and $b_n = \frac{1}{n+1}$, then they become arbitrarily close but $|f(a_n) - f(b_n)| = 1 \not\to 0$. So $f$ is not uniformly continuous.
Example 2. Let $f(x) = x^2$ on $\mathbb{R}$. If we pick two 'infinitely large' $a_n = n$ and $b_n = n+\frac{1}{n}$, then $a_n$ and $b_n$ become arbitrarily close, but $|f(a_n) - f(b_n)| = 2 + \frac{1}{n} \not\to 0$. So $f$ is not uniformly continuous.