A vector space is a set together with operations called addition and scalar multiplication that satisfies some rules (distributive property, etc.). One of the rules is that there must be an identity element for addition, and this is called (conventionally) "$0$".
An affine space is a pair $(A, W)$, where $A$ is a set and $W$ a vector space, and an operation $T: A \times W \to A$ (called "translation") with certain properties, like
- $T(a, 0) = a$ for all $a$ ("translation by $0$ leaves things fixed")
- $T(a, v+w) = T(T(a, v), w)$ ("translations add nicely")
- If $T(a, v) = a$ for some $a \in A$, then $v = 0$ ("no degenerate translations except translation by $0$")
The claim being made in that Wikipedia article is that if you have a vector space $V$, there's an easy way to build an affine space $S$, which is this:
Let $A$ be the set of all the vectors in $V$, considered merely as a set (i.e., forgetting that there are addition and scalar multiplication operations on $A$).
Let $W = V$.
Define $T(a, v) = a + v$, where the addition on the right hand side is "the addition operation on the vector space $v$.
This set of three things, $A, V, T$, constitutes an affine space.
Note that if you have an affine space, there's a way to get a vector space from it, namely, from the pair $(A, V)$, simply take $V$.
These two operations are "partial inverses" in the sense that if you perform the first and then the second, you get back your original vector space. If you perform the second and then the first, you don't get back the same affine space, but you do get back one that's "affine equivalent" to the one you started with (which requires a notion of equivalence, which takes more writing than I'm willing to do, given that this isn't even part of your question).
To give explicit answers:
In a vector space, the vector called $0$ has a special role in that it's the identity element for the addition operation.
It'd perhaps be better, in your initial understanding, if you associated to each pair of points in the underlying set of $V$ an actual MAP, i.e., a translation from the set $V$ to itself. That'd look like this:
V \times V \to Trans(V) : (v, w) \mapsto (u \mapsto u + v - w).
In this context, you can see that the pair $(0,0)$ maps to the "don't move anything" translation, but so does $(v, v)$ for any vector $v$, and in that sense, the $0$ vector is nothing special.