One can interpret a homomorphic image of an algebraic structure as a "collapsed" or "low-resolution" version of it, since different elements of the original structure get blurred together into becoming the same pixel in the image. Thus if we have a chain of surjective homomorphisms, we are getting higher and higher resolution pictures of ... something, "in the limit." That something is the inverse limit.
Consider real numbers for a moment. What are they? Most people have an intuitive idea of a continuum and maybe have been exposed to the phrase "number line," but only a fraction - people who study math - get to know how we actually construct real numbers. People can cite decimal expansions, for instance that of pi, $\pi=3.1415\dots$, but what does that mean? It means there is a series of approximations of indefinitely increasing precision, $3$, $3.1$, $3.14$, $3.141$, $3.1415$, $\cdots$. In fact one formal construction of the real numbers is to identify a real number with a(n equivalence class) of Cauchy sequences. We can think of these Cauchy sequences (of rationals) as sequences of approximations of "ghosts" that are not really there in the set of rationals.
It is similar with inverse limits. Think about $\Bbb Z/p\Bbb Z\leftarrow\Bbb Z/p^2\Bbb Z\leftarrow\Bbb Z/p^3\Bbb Z\leftarrow\Bbb Z/p^4\Bbb Z\leftarrow\cdots$ with the canonical projection maps. Every positive integer can be written as $a_0+a_1p+\cdots+a_rp^r$, i.e. can be written in base $p$. What does the projection map $\Bbb Z/p^{r+1}\Bbb Z\to\Bbb Z/p^r\Bbb Z$ do to the digital representation of an integer residue? Simple: it simply deletes the $p^r$ digit. Thus, any "sequence of approximations" can be identified with an infinite base $p$ expansion, $a_0+a_1p+a_2p^2+\cdots$ (the sequence of approximations tells us what every digit in the infinite expansion is, and vice-versa).
In general, given a chain $G_0\leftarrow G_1\leftarrow G_2\leftarrow\cdots$, and a sequence $(g_0,g_1,g_2,\cdots)$ which is appropriately "compatible" (i.e. $g_{i+1}\mapsto g_i$) we can think of it as a "convergent sequence" whose limit is an element of the inverse limit $\varprojlim G_i$. The operation(s) can be done pointwise. This construction works just as fine if we have a nonlinear system of surjective homomorphisms.
Now let's think about Galois automorphisms. Given any tower $L/M/K$ in which all three extensions are Galois we know there is a surjection ${\rm Gal}(L/K)\to{\rm Gal}(M/K)$. One can consider the entire system of such projection maps for algebraic extensions of a given field $K$, and then consider $\varprojlim{\rm Gal}(L/K)$ (note that the groups are varying with the field $L$ in this limit). What does an element $\sigma$ of this inverse limit look like, tangibly? Well, for any Galois extension, there is a corresponding automorphism $\sigma|_L$ of the extension $L/K$, so ultimately $\sigma$, via all of these $\sigma|_L$s, "knows" where to send every element which is algebraic over $K$, which means this element "knows" of an automorphism of $\overline{K}/K$. And vice-versa, since any element of ${\rm Gal}(\overline{K}/K)$ restricts to an automorphism of any Galois extension $L/K$.
If we view $\varprojlim{\rm Gal}(L/K)$ as the subgroup of $\prod_L{\rm Gal}(L/K)$ comprised of "compatible" elements (recall, that means our "convergent sequences"), then the obvious map ${\rm Gal}(\overline{K}/K)\to\varprojlim{\rm Gal}(L/K)$ (where the "$L$th" coordinate of where $\sigma$ is sent to is simply the restriction of $\sigma$ to $L$) is a group isomorphism. The corresponding inverse map is simply the process of patching together all of the compatible automorphisms of Galois extensions of $K$ into one giant automorphism of $\overline{K}/K$.
One can also endow these inverse limits with the profinite topology in which case we have an isomorphism of topological groups. Another way to define the inverse limit is via universal properties in category theory. Uniqueness of inverse limits (up to unique isomorphisms) follows from a simple abstract nonsense argument, and existence follows from using the explicit construction I discussed above.
Let me try to explain my understanding of what Abel and Ruffini did. I am thinking of using this as my opening lecture next time I teach Galois theory, so this is useful for me. Disclaimer: I am a mathematician, not a historian.
Here is a theorem which I can prove in the space on this page.
Theorem. Let $L= \mathbb{C}(x_1,\ldots, x_5)$ be the ring of rational functions
in $5$ variables, and let $K$ be the subfield of symmetric functions.
So $K$ is generated by $e_1 = x_1+x_2+\cdots+x_5$, $e_2=x_1 x_2 + x_1 x_3 + \cdots + x_4 x_5$, ... and $e_5 = x_1 x_2 \cdots x_5$. Suppose
we add elements to $K$ by the operations of $+$, $-$, $\times$, $\div$
and $\sqrt[n]{}$, while staying within $L$. Then we can never
reach the element $x_1 \in L$.
Proof: The group $S_5$ clearly acts on $L$ by permuting the $x_i$. Let $\sigma$ and $\tau$ be the permutations $(123)$ and $(345)$, and let $F$ be the field of functions in $L$ fixed by $\sigma$ and $\tau$. Clearly, $K \subset F$. We claim that you cannot escape $F$ by the operations $+$, $-$, $\times$, $\div$, $\sqrt[n]{}$. For the first four operations, this is obvious.
Suppose that $b \in L$ and $b^n=a \in F$. Then $\sigma(b)^n=\sigma(b^n) = \sigma(a) = a=b^n$. So $(\sigma(b)/b)^n=1$ and $\sigma(b) = \zeta b$ for some $n$-th root of unity $\zeta$. Similarly, $\tau(b) = \omega b$ for some $n$-th root of unity $\omega$.
Now, $\sigma^3=\mathrm{Id}$ so $b=\sigma^3(b)=\zeta^3 b$. Similarly, we deduce that $\omega^3=1$. Also, $(\sigma \tau)^5=\mathrm{Id}$, so $\zeta^5 \omega^5=1$, and $(\sigma^2 \tau)^5 = \mathrm{Id}$ so $\zeta^{10} \omega^5=1$. Combining all of these equations, $(\zeta^3)^2 (\zeta^5 \omega^5) (\zeta^{10} \omega^5)^{-1} = \zeta =1$ and, similarly, $\omega=1$. So $\sigma(b) = \tau(b) = b$, and $b$ is in $F$ after all.
Since $x_1 \not \in F$, we have proved the theorem $\square$.
Note that the quadratic, cubic and biquadratic formulas do stay within $L$.
For example, the cubic formula is (Double check before using!)
$$x_1 = \frac{1}{3} \left( \sqrt[3]{\frac{S+\sqrt{S^2+4 T^3}}{2}} + \sqrt[3]{\frac{S-\sqrt{S^2+4 T^3}}{2}} + e_1 \right)$$
$$S = 2 e_1^3-9 e_1 e_2+27 e_3, \quad T= 3e_2-e_1^2.$$
The functions $S$ and $T$ are formed by field operations and I believe you will find that $\sqrt{S^2+4T^3} = (x_1-x_2)(x_1-x_3)(x_2-x_3)$ and $$\sqrt[3]{\frac{S+\sqrt{S^2+4T^3}}{2}} = x_1 + \frac{-1+\sqrt{3} i}{2} x_2 + \frac{-1-\sqrt{3}i}{2} x_3$$ So we have stayed within $L$ while working our way up to $x_1$.
I explain how to think about this computation in modern terms here. In brief, the group generated by $\sigma$ and $\tau$ is $A_5$, and the above computations verify that $A_5^{ab}$ is trivial.
As I understand it, both Abel and Ruffini gave correct proofs of the theorem in the box. These proofs were basically the same as the above, but longer because words like "field", "group" and "action" hadn't been invented yet. Both of them aimed to go further, and show that there was no universal formula for $x_1$ in terms of $e_1$, $e_2$, ..., $e_5$, $+$, $-$, $\times$, $\div$ and $\sqrt[n]{}$, whether or not we are required to stay in $L$. This was very confusing, as it wasn't clear what sort of algebraic object the formulas would live in once we left $L$.
As I understand it, the current consensus is that Ruffini's attempt failed, but Abel succeeded by proving the result now called the Theorem of Natural Irrationalities.
From a modern Galois theory perspective, there is no problem. Suppose we have some chain of fields $K \subset F \subset L \subset M$, and $K=K_0 \subset K_1 \subset K_2 \subset \cdots \subset K_r = M$, where each $K_{i+1}/K_i$ is a radical extension. Without loss of generality, we may assume $M/K$ is Galois, with Galois group $G$. Then the chain of radical extensions shows $G$ is solvable. But the chain $K \subset F \subset L \subset M$ shows that $A_5 = \mathrm{Gal}(L/F)=\mathrm{Gal}(M/F)/\mathrm{Gal}(M/L)$ occurs as a composition factor for $G$, so $A_5$ must also be solvable. This contradicts that our computation that $A_5^{ab}$ is trivial.
What Galois contributed was the ability to work with arbitrary field extensions, not just polynomials/functions with various symmetry, and therefore be able to speak cleanly about the field $L$ and its symmetries. He also was the first to create tools that would let us prove a particular quintic over $\mathbb{Q}$ was not solvable by radicals.
Best Answer
I suggest that you read Niels Hendrik Abel and Equations of the Fifth Degree by Michael Rosen for a proof of the Abel-Ruffini theorem.