Need help understanding the relation between Galois theory and a general quintic formula impossibility.

Firstly, when we talk about the Galois group of some polynomial $f \in F[x]$, we are referring to the Galois group of the splitting field $K$ of $f$ over $F$.

Now if there was a formula for the roots only using $n^{th}$ roots, then $K$ would be a radical extension of $F$, by which I mean there is a (necessaily finite) chain of fields $F=F_0 \subset F_1 \subset \cdots \subset F_n=K$, where $F_i=F_{i-1}(\sqrt[n_i]{\alpha_i})$ for some integer $n_i$, and $\alpha_i \in F_{i-1}$.

Now it can be shown that radical Galois extensions have soluble Galois groups. As an outline, the idea of the proof for this comes from the Galois correspondence and a result form group theory that if $N \triangleleft G$ is a normal subgroup and both $N$ and $G/N$ are soluble, then so is $G$.

So now let's assume that $f$ is soluble by radicals. Then its splitting field is a radical extension and therefore the associated Galois group must be soluble by the above. Hence if we discover that the Galois group is for example $S_5$, i.e. something insoluble, then we must be mistaken in our original assumption that $f$ was soluble by radicals.

The only thing left to finish this is to exhibit such a polynomial, since a priori it may happen that all Galois groups have to be soluble (this does actually happen for certain types of fields). Such an example is $f=x^5-4x-1 \in \mathbb{Q}[x]$.

To answer the last point, there is nothing particularly special about $S_5$; it is just the standard example to choose since we need at least a degree $5$ polynomial to construct an insoluble Galois group and the problem of finding polynomials whose Galois groups are isomorphic to a symmetric group was one of the first cases of inverse Galois theory to be done.