Uniform Convergence: The two definitions I know of the two equivalent definitions for uniform convergence. Namely:
$f_n(x)$ converges uniformly to $f(x)$ if either:
a)$$ \forall \epsilon \exists N \in \mathbb N s.t. \forall x \in D \ \forall n \geq N:
|f_n(x) - f| < \epsilon $$
or
b) $$\lim_{n \rightarrow \infty} \sup_{x\in D} |f_n(x) - f(x)| = 0 $$
While I see why in the case of a) the same "speed of convergence" is guaranteed since one epsilon is chosen for all $x$, I unfortunately cannot make sense of b) on an intuitive level. So I see that b) states that the biggest difference between the sequences of the functions and the limit functions must become zero for n going to infinity. But how does that also guarantee this convergence process is of "uniform speed" for all x?

If I understand correctly, you want to convince yourself of (b)$\to$(a).
Assume (b), and let $\epsilon>0$ be an arbitrary positive number. Since we assumed that $\sup_{x\in D}|f_n(x)-f(x)|\to0$, there exists $N$ such that $\sup_{x\in D}|f_n(x)-f(x)|<\epsilon$ for all $n\ge N$, that is, for all $x\in D$, and for all $n\ge N$, $|f_n(x)-f(x)|<\epsilon$, that is, (a).
The $\sup$ is there just as a short way of saying that all $|f_n(x)-f(x)|$ are smaller than something - if the supremum of a set is lower than $\epsilon$, then clearly every element in that set is lower than $\epsilon$, and the limit is just a short way to write the whole $\forall \epsilon \exists N$ part.
