How is the Toda bracket affected by choice of nullhomotopy?

Regarding your first question, the Toda bracket $\langle h, g, f \rangle$ indeed "only" depends on the homotopy classes of $f$, $g$, and $h$.

Here is a somewhat pedantic answer to your second question. The Toda bracket should be defined as a subset of $[\Sigma W, Z]$ consisting of all the maps you can get from the construction described by Hatcher as you vary over all the nullhomotopies of $gf$ and $hg$. Therefore, I would say that the Toda bracket (as a subset) trivially does not depend on the choices of nullhomotopies.

Of course, Hatcher is using "Toda bracket" in a slightly different way, to refer to a particular map you get after selecting specific choices of nullhomotopies. If you're wondering about how to go about showing the exercise Hatcher gives about the indeterminacy, here is a sketch of the argument. Consider the set of nullhomotopies of $gf$. It is a torsor for $[\Sigma W, Y]$. To see this, think of a nullhomotopy as a map $CW \to Y$, and if you have two such maps you can glue the cones along $W$ to get a map $\Sigma W \to Y$. Conversely, given a nullhomotopy $CW \to Y$ and a map $\Sigma W \to Y$, you can glue one of the double cones in $\Sigma W$ to cancel out $CW$, and you're left with a new map $CW \to Y$. Therefore, one source of indeterminacy comes from $h_* [\Sigma W, Y]$. Arguing analogously for the nullhomotopy of $hg$, we have another indeterminacy coming from $f^* [\Sigma X, Z]$. Therefore altogether the indeterminacy is given by the subgroup $f^* [\Sigma X, Z] + h_* [\Sigma W, Y] \subseteq [\Sigma W, Z]$, and the Toda bracket is a coset of this subgroup.