I am trying to convert the Bessel equation $$z^2u''+zu'+(k^2z^2-v^2)u=0,$$ into the Sturm-Liouville form.
Dividing by $z$ ($z\neq 0$), we yield $$zu''+u'+(k^2z-v^2z^{-1})u=0\implies (zu')'+(k^2z-v^2z^{-1})u=0.$$ Now I believe this is in Sturm-Liouville form, as it is of the form $$(p(z)u')'+(q(z)+\lambda r(z))u=0.$$ My question is, how do we know what values $q(z)$ and $r(z)$ correspond to?
The parameter $v$ is something that comes from separation of variables in other variables. So it should be treated as fixed. Your parameter $k^2$ is typically the eigenvalue parameter associated with the Bessel equation. So $\lambda=k^2$. The Sturm-Liouville form for the Bessel equation for a fixed $v$ is $$ (zu')'+(-v^2/z+k^2z)u=0 $$ which matches your form
$$ (pu')'+(q+\lambda r)u = 0, \\ p(z)=z,\;\; q(z)=-v^2/z,\;\; r(z)=z^2,\;\; \lambda=k^2. $$