What's so special about square root cancellation?
Solution 1:
Each of these statements are of the form "this sum has size consistent with the terms being independently uniformly randomly distributed". Many number theoretic objects are like this --- they have lots of multiplicative structure but also behave in many ways like random variables.
It isn't true, of course, that the primes are independent random variables. But it is interesting that various quantities related to them seem to behave like random variables.
The reference idea is a random walk. You walk on a $d$-dimensional lattice. At each step, you choose a random direction in the lattice and walk one step in that direction. After $N$ steps, the expected absolute distance from the starting place is approximately $\sqrt{\tfrac{2N}{d}} \Gamma(\tfrac{d+1}{2})/\Gamma(\tfrac{d}{2})$ (and approaches this quantity as $N \to \infty$). This is the same as asking about the expected magnitude of a sum of $N$ identical independent random variables whose values are uniform across the unit in the associated lattice. Then what this result on random walks tells us is that on average, we expect the magnitude of these random sums to be of size roughly $\sqrt{N}$ (times a constant), and thus to exhibit a form of square root cancellation.
Similar statements can be stated more generally and there are many proofs. As the comments allude to, these random walk results can be proven by appealing to the central limit theorem in probability.
In your analytic number theory note, you have evidently shown that $\sum_{n \leq x} n^{-it} = O(\sqrt{x})$, and thus also exhibits square root cancellation --- which is what you would expect if the $n^{-it}$ were identical independent random variables.
I'll note that the specific quote you translate [about how RH implies that the primes are distributed as uniformly as possible] is morally true, but at a rather high level. I would not dwell too much on it as it makes sense mostly in the context of the analytic proof of the prime number theorem.