Need help understanding the relation between Galois theory and a general quintic formula impossibility. Recently I have been studying a fair amount of Galois Theory and think I understand a lot of the ideas and theorems (at least intuitively) and how the things kind of come together and how Galois would have had motivation to look at certain topics. 
In particular I just finished looking at the Galois Correspondence theorem which provides us with a way to "transform" groups into fields and use one to understand problems in another.
More specifically we have a correspondence between subgroups of $\text{Gal}(E/F)$ and intermediate fields of the Galois extension $E/F$. 
I want to understand (at least at a basic/intuitive level (I am not interested in getting bogged down in proofs yet I just want to straighten out the concepts involve and how they intertwine)) how this means that there is not a general forumla (using addition, subtraction, multiplication and nth roots) for polynomials of degree $5$ (I understand this can be extended to higher degrees but I'm not interested in that yet either.)
I understand the reason for this is that there are $5$th degree polynomials whose Galois group is the whole of $S_5$ which is not a solvable group (since $A_5$ is not solvable and $A_5$ is the only intermediate normal subgroup of $S_5$). And we have a polynomial is solvable if and only if its Galois group is a solvable group. Hence for this reason we cannot find a general formula for quintics. 
There are several parts I don't understand about this and would be very appreciative if someone could clear these issues up for me so I can get more of an idea as to what is going on (as said I am less interested in proofs and rigour at the moment and want to get a feel for what is going on). 
I will try and summarise my main queries below:
1) What exactly is the relationship between the Galois group of a polynomial (the set of permutations of roots that will still satisfy any given polynomial that it satisfied before) and the $\text{Gal}(E/F)$? The correspondence talks about subgroups of $\text{Gal}(E/F)$ and subfields of $E/F$ and I understand we want to use the correspondence to switch from looking at a problem in fields to looking at the corresponding problem in group theory. But the relationship doesn't talk about the Galois group of a polynomial (presumably it does and the two ideas are linked I just don't see how).
2) How exactly is the Galois correspondence used to show there isn't forumlas for quintics? I outlined the rough argument as I understand it but it is not clear at all to me when it is applied.
3) When we talk about quintics having no formula are we talking about quintics with rational coefficients, real coefficients, complex coefficients or all of them?
4) What exactly is meant by the definition of the Galois group of a polynomial?
(We can shift roots and some equations won't notice we've done anything) which equations are we talking about? I see $x^2-2$ can't tell if we switch $\sqrt2 \mapsto -\sqrt2$ for instance.
5) We want to show there is a polynomial whose Galois group is $S_5$. To me this means we want to find a polynomial with roots that we can move around as much as possible. I know there is a theorem that says there is no formula (i.e. solutions in radicals) if the Galois group is not solvable like $S_5$ but why is this the case what is so special about the fact we can move around the roots so much and nothing is effected (presumably this has something to do with the Galois correspondence)?
I think that is all I can think of asking for now.
I have tried my best to make it clear as to where my understanding is letting me down.
It's like I have all these pieces of information that I understand separately but I'm missing something very very important to see the bigger picture and bring everything together. (I kind of understand most of the Galois theory I have learned but I don't know how to apply it to the quintic problem basically.)
I would very much appreciate if someone could explain in as basic a way as possible exactly what I am missing to tie this all together.
Thanks for reading I appreciate your time.
I know this post is long but I wanted to lay out all my thoughts to try and make it as easy for people to help me as possible. If you want to just answer parts that is fine but if someone would be able to write something that tries to clarify it all for me then I am patient and happy to wait.
I look forward to your replies :)
 A: 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.
A: Very good attempt and I would say gaining intuition is extremely important as much as understanding the proofs.
I am in the same boat where you are and this is my understanding. There is a leap of concepts when we move from Galois correspondence to solvability using radicals. When we talk about Galois group, we are talking about the group associated with a single polynomial $f$ and the splitting field of that polynomial. 
Any intermediate normal extension lying between the base field and the splitting field of the original polynomial $f$ is essentially a splitting field of a different irreducible polynomial over the base field. 
In other words, there cannot be a chain containing two normal extensions for the same polynomial, since normal extensions for a polynomial, by definition, should contain all of its roots unless they are irreducible over the field.
However, when we talk about solvability, we are considering a set of polynomials solvable through radicals. Here cyclotomic polynomials and their roots serve as guiding factor since those polynomials are solvable through radicals and they give a blueprint for the general solution of polynomials through arithmetic operations and radicals. 
So a chain of normal extensions for a solvable polynomial is not for a single polynomial you pick, (for example, let's say you pick $x^7-7$, this will have just one normal extension corresponding to this polynomial; the extension field adjoining only the seventh primitive root of unity will correspond to a different irreducible polynomial), but for any generic solvable polynomial like $x^k-a$ for which the chain would apply. There is a kinda meta-logic here which needs careful thought. 
