The existence of partial fraction decompositions

Another way of looking at Arturo's comments about unique factorization: the key to the standard partial-fraction decomposition is the existence of the Euclidean Algorithm for polynomials, along with the standard-but-unspoken assumption that $g(x)$ and $h(x)$ in your initial decomposition have no factor in common - that is, that their GCD is 1. If we can find a partial-fraction decomposition for $1\over g(x)h(x)$ as ${A(x)\over g(x)}+{B(x)\over h(x)}$, then we can certainly find one for $f(x)\over g(x)h(x)$ for any polynomial $f$; just multiply by $f(x)$ in that decomposition. Likewise, if we can find a partial-fraction decomposition for $f(x)\over g(x)h(x)$ for any polynomial $f$, then we can certainly find one for the special case $f(x)\equiv 1$. But saying that ${1\over g(x) h(x)} = {A(x)\over g(x)}+{B(x)\over h(x)}$ is the same as saying that ${A(x)h(x) + B(x)g(x) \over g(x)h(x)} = {1\over g(x)h(x)}$, or in other words saying that $A(x)h(x) + B(x)g(x) \equiv 1$, and the existence of polynomials $A$ and $B$ with these properties is exactly what the (extended) Euclidean algorithm provides. It doesn't work for your case of $f(x)\over g^2(x)$ because $GCD(g(x),g(x)) \neq 1$, so there can't be any $A(x)$ and $B(x)$ satisfying $A(x)g(x)+B(x)g(x) \equiv 1$.

No, in general you cannot split it that way. For example, $$\frac{x}{(x+1)^2}$$ cannot be written as $$\frac{A}{x+1} + \frac{B}{x+1}$$ with $A$ and $B$ constants. Because you would just get $$\frac{A}{x+1} + \frac{B}{x+1} = \frac{A+B}{x+1},$$ with $A+B$ constant, so this is definitely not equal to $\frac{x}{(x+1)^2}$.

If you have $$\frac{A}{g(x)} + \frac{B}{g(x)}$$ then the answer is just $$\frac{A+B}{g(x)},$$ just like with fractions, you have $$\frac{A}{n} + \frac{B}{n} = \frac{A+B}{n}.$$ So unless $\frac{f(x)}{(g(x))^2}$ can be simplified by cancelling one of the factors of $g(x)$, you have no hope of writing it as a sum of two fractions with $g(x)$ in the denominator.

(Somewhat hidden in the above is the fact that polynomials also have unique factorization, so if $$\frac{f(x)}{g(x)} = \frac{h(x)}{k(x)}$$ then $f(x)k(x) = g(x)h(x)$. If $f(x)$ and $g(x)$ have no common factors and $h(x)$ and $k(x)$ have no common factors, then the equality means that you must have $g(x)=k(x)$ and $f(x) = h(x)$ up to multiplication by constants; again, just like an equality of fractions $$\frac{a}{b} = \frac{c}{d}$$ with $a,b,c,d$ all integers, and $a$ and $b$ have no common factors and $c$ and $d$ have no common factors, then up to sign you must have $a=c$ and $b=d$.)

HINT $\rm\displaystyle\qquad \frac{1}{g\:h}\ =\ \frac{a}g\ +\ \frac{b}h\ \iff\ 1\ =\ a\ h\: +\: b\ g\ \iff\ 1\: =\: \gcd(g,h)$