Counterexample to if g ◦ f is surjective, then f and g are surjective I want less of an answer and more of an explanation if possible please.
I understand I'm looking for a range of either f and g that is NOT surjective(does not cover all the codomain), but that their composition is. 

With questions like this, it is often good to look for the simplest possible example. The one (to me!) that comes to mind is the following.
Let $X = \{0\}$ and let $Y = \{0, 1\}$ (realistically, $Y$ can be any set with more than one element). Define the following functions:
$$
f : X \to Y \qquad \qquad 0 \mapsto 0
$$
and
$$
g : Y \to X \qquad \qquad 0, 1 \mapsto 0
$$
These are both functions, and we note first of all that $f$ is not surjective. However, if we look at the composition $g \circ f$ then it takes $0 \mapsto_f 0 \mapsto_g 0$ and so this composition is just the identity function on $X$, and so is surjective. So we have found a pair of functions $f, g$ such that their composition is surjective, but that at least one of the two functions is not.
Again, I think it is helpful to look at the simplest possible examples. Try sets with one or two elements. Draw pictures, if need by (the usual blob-with-arrow version of functions). Play around!
