Closed form formulas are overrated. When they exist, generating function techniques can often help you find them; when they don't, the generating function is the next best thing, and it turns out to be much more powerful than it looks at first glance. For example, most generating functions are actually meromorphic functions, and this means that one can deduce asymptotic information about a sequence from the locations of the poles of its generating function. This is, for example, how one deduces the asymptotic of the partition numbers.
In your particular example, the generating function is rational, so it has a finite number of poles. That means you can use partial fraction decomposition on it to immediately get a closed form.
You might be interested in reading my notes on generating functions, which have several examples and which I hope will be enlightening. The first basic thing to grasp is that manipulating generating functions is much easier than manipulating sequences, but the power of generating functions goes much deeper than this. For a really thorough discussion I highly recommend Flajolet and Sedgewick's Analytic Combinatorics, which is available for free online.
To quote the introductory words of Wilf in Chapter 1 of his excellent book "generatingfunctionology",
A generating function is a clothesline on which we hang up a sequence of numbers for display.
Specifically, to answer your question on what it means to talk of the "generating function of a sequence": Given a sequence $a_0, a_1, a_2, \dots$, the generating function of that sequence is the object $A(x) = a_0 + a_1x + a_2x^2 + \dots$. This is a way of encoding all the elements of the sequence into a single object.
For instance, consider the very simple sequence defined by $a_n = 1$ for all $n$. Its generating function is the object $A(x) = 1 + x + x^2 + \dots$, which can also (if you wish) be written more concisely as $\dfrac{1}{1-x}$. Or, consider the sequence defined by $a_n = 2^n$. The generating function of this sequence is the object $A(x) = 1 + 2x + 2^2x^2 + 2^3x^3 + \dots$, which can also be written as $\dfrac{1}{1-2x}$. Note that this gives a way to encode all the (infinitely many) elements of the sequence $1, 2, 2^2, 2^3, \dots$ into a single object $A(x)$. (Of course there are other ways too: you could say that the function $n \mapsto 2^n$ is a single object.)
Anyway, now that you hopefully understand roughly what generating functions are, we can turn to their use in counting problems. In general, when we want to find an element of some sequence, one way is to find the generating function of the whole sequence first, and then find the particular element we care about. This may seem like more work, but in practice can often be simpler.
So, let $a_n$ be the number of solutions to $3a + 4b + 2c + d = n$; we want to find the generating function $A(x) = a_0 + a_1x + a_2x^2 + \dots$ and then read off $a_{20}$ which is the coefficient of $x^{20}$ in it. All this should answer what generating functions are here, why take coefficient of $x^{20}$ etc.
The actual part related to the counting problem is relatively simple (though nontrivial) once you understand all this: if you consider $(1 + x^3 + x^6 + \dots)$, the coefficient of $x^k$ in this is $1$ if $k$ can be written as $3a$ for some $a$, and $0$ otherwise. Similarly the other factors. For any solution $(a, b, c, d)$ satisfying $3a + 4b + 2c + d = n$, when you multiply the four factors $(1+x^3+x^6+\dots)(1+x^4+x^8+\dots)(1+x^2+x^4+\dots)(1+x+x^2+\dots)$ together, you'll get a bunch of terms, where each term in the product is got by taking one particular term from each of the factors, and multipying those four terms together. One of the particular products is that of taking $x^{3a}$ from the first factor, $x^{4b}$ from the second, etc., to give $x^{3a}x^{4b}x^{2c}x^{d} = x^{3a + 4b + 2c + d} = x^n$. Each solution $(a, b, c, d)$ contribues one such term $x^n$, and these solutions are the only way to get $x^n$ in the product. So the coefficient of $x^n$ is the number of times it occurs in the product, which is the number of solutions.
Best Answer
You are looking for the coefficient of $x^{24}$ in:$$x^{12}\left(\frac{1-x^6}{1-x}\right)^4$$
Which is the same as the coefficient of $x^{24-12}=x^{12}$ in $$\left(\frac{1-x^6}{1-x}\right)^4\tag{1}$$
Now, $$(1-x^6)^4 = 1-4x^6+6x^{12}-4x^{18}-x^{24}$$
And $$\frac{1}{(1-x)^4} = \sum_{n=0}^{\infty} \binom{n+3}{3}x^n$$
So what is the coefficient of $x^{12}$ in the product? It is $$\binom{12+3}{3}\cdot 1 + \binom{6+3}{3}\cdot (-4) + \binom{3}{3}\cdot 6$$
More generally, the coefficient of $x^k$ in (1) is:
$$\binom{k+3}{3}\cdot 1 + \binom{k-3}{3}\cdot (-4) + \binom{k-9}{3}\cdot 6 +\binom{k-15}{3}\cdot (-4) + \binom{k-21}{3} \cdot 1$$
And yes, if $k>20$ this will evaluate as zero.