To answer the second question first: You can certainly apply the product rule to each multiplication in turn. However, since multiplication is commutative and associative, the result will necessarily be symmetric with respect to the three factors, so it makes sense to derive a general rule and remember it instead of going through the factors pairwise in each case.
Applying the product rule to the product $fgh$ by grouping it as $(fg)h$ yields
(fgh)'=((fg)h)'= (fg)'h + (fg)h' = (f'g + fg')h + (fg)h'=f'gh+fg'h+fgh'\;.
Perhaps you can guess from this result what the general result for an arbitrary number of factors will be:
\left(\prod_i f_i\right)'=\sum_if_i'\prod_{j\ne i} f_j\;.
You can either prove this using induction (which might be a good exercise), or you can see why it's true by considering what happens if you change each of the factors a bit:
$(f+\Delta f)(g+\Delta g)(h+\Delta h)\cdots=fgh\cdots+\Delta fgh\cdots+f\Delta gh\cdots+fg\Delta h\cdots+\cdots\;,$
simply from multiplying out the parentheses. The remaining terms are at least of second order (i.e. they contain at least two $\Delta$ terms), so they don't end up in the derivative, which gives the first-order approximation to the function.
Turning now to your first question, yes, you can apply the product rule instead of the quotient rule; well spotted. In fact, the quotient rule is nothing but a convenient shortcut for applying the product rule:
\left(\frac fg\right)'=\left(fg^{-1}\right)'=f'g^{-1}+f\left(g^{-1}\right)'=f'g^{-1}+f\left(-\frac{g'}{g^2}\right)=\frac{f'g-fg'}{g^2}\;.