This is a tentative answer requested by Tito.
Starting aroundin 2008 I worked on a collection of Ramanujan theta function identitiesfunction identities as an extension of my Dedekind eta product identities collection. In order to do this effectively, I needed to know theto know what power of $q$ factor to use which "completes" these functions functions. My source source was Bruce Berndt, Ramanujan Notebooks, Part Part III, page 42 page $42$ which states
$ G(q) = q^{(m-n)^2/8(m+n)}f(q^m,q^n) .$
For example, forif $f(-q^k) := f(-q^k,-q^{2k}),$ the$q=e^{2\pi i\tau}$, then $\eta(k\tau) = q^{k/24}f(-q^k) = q^{k/24}f(-q^k,-q^{2k}),$ and the power of $q$ factor is thus $q^{k/24}$.
For another example, $$h_1 = \frac{q^{121/104}f(-q,-q^{12})}{q^{5/24}f(-q^5)} = q^{149/156}\frac{f(-q,-q^{12})}{f(-q^5)} = \\ q^{149/156}(1 - q + q^5 + 2q^{10} - 2q^{11} - q^{12} + 4q^{15} +\dots) .$$
Another source for $p=13$ is R. J. Evans, Theta Functions Identities, $1990$, pages 97, 99, 113, 116$97, 99, 113, 116$.
(By TP). As allowed by the OP, the formula for the exponent $d$ of the $q$-factor
$$q^d\, f(q^m,q^n)$$
for level $13$ seems to be,
$$d(m,n) = \frac{(m-n)^2}{8(m+n)}-5/24$$
This yields,
$$d(1,12) = \frac{149}{156}, \quad d(2,11) = \frac{89}{156}, \quad d(3,10) = \frac{41}{156}$$ $$\quad d(4,9) = \frac{5}{156},\quad d(5,8) = -\frac{19}{156}, \quad d(6,7) = -\frac{31}{156}$$
This leads to the ratio,
$$r_2 = \frac{h_4}{h_2} = \frac{q^{5/156}f(-q^4,-q^{9})}{q^{89/156}f(-q^2,-q^{11})} = q^{-7/13}\frac{f(-q^4,-q^{9})}{f(-q^2,-q^{11})}$$
and the reduced $q$ factor matches the one in Ramanujan's ratio formula. This implies the $q$ factors in the addendum are in error because the wrong pairings were used.
