Why are generating functions useful?

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.

Have you read Generatingfunctionology by Herbert Wilf? The book is loaded with methods and techniques based on generating functions for solving a number of different problems. I guess the book can answer your question better than I could.

I feel that generating functions are powerful because they allow you to use tools from calculus and analysis to solve problems in areas such as discrete mathematics and combinatorics, where such tools don't seem readily applicable.

Edit: I just wanted to add one more point. Suppose you have translated the problem from the sequence to the generating function and you don't see any simple closed form solutions, you could use Wolfram Alpha to make absolutely sure. Wolfram alpha can easily expand out expressions involving large polynomial terms and will often give you a series expansion if it exists in closed form.

To answer your question for the example you state:

$$ \frac{1}{1-x^2} \cdot \frac{x}{1-x^2} \cdot \frac{1-x^4}{1-x} = (x+x^2 +x^3 + x^4)(1-x^2)^{-2} $$

By binomial theorem (yes, it is valid for negative exponents too) we get that this is equal to

$$ (x + x^2 + x^3 + x^4) \left(\sum_{j=0}^{\infty} (-1)^j {-2 \choose j} x^{2j}\right)$$

Can you now figure out a formula for the coefficient of $\displaystyle x^n$, thus solving your problem?