Which element has a greater second electron affinity, sulfur or oxygen?
The most important fact here is that the electron added to $\ce{S-}$ goes into an $n=3$  orbital ($3p$), whereas the one added to $\ce{O-}$ goes into an $n=2$ orbital ($2p$).

The larger $n=3$ orbitals are further from the nucleus and hence experience a weaker attraction to the nucleus and are much higher in energy.  This dependence with quantum number scales approximately $E\sim - n^{-2}$ so is very important [1].

The dependance of orbital energy also scales quadratically with effective nuclear charge, however effective nuclear charges falls significantly compared to the true nuclear charge due to shielding and repulsion from other electrons.  A common first order approximation for the effective nuclear charge is given by Slater's rules [2]

Combining these two approximations we can calculate very approximate orbital energies.

$$
\begin{align}E_{2p}(\ce{O^2+}) &amp; {} = - R_\infty h c \frac{Z_{eff}^2}{n^2} \\
&amp; {} = - R_\infty h c \frac{(8-4.15)^2}{2^2} \\
&amp; {} = - 3.7 \text{ Ry} \\
E_{3p}(\ce{S^2+}) &amp; {} = - R_\infty h c \frac{Z_{eff}^2}{n^2} \\
&amp; {} = - R_\infty h c \frac{(16-11.25)^2}{3^2} \\
&amp; {} = - 2.5 \text{ Ry}
\end{align}
$$

Where 1 Ry is the electronic energy of a hydrogen atom.

As you can see, to first order, the energy of the new electron is much lower when added to $\ce{O-}$ than when added to $\ce{S-}$, leading to a more favourable electron affinity for $\ce{O-}$ than $\ce{S-}$, driven by the fact that the orbital occupied has a lower principal quantum number $n$

[1] - See the energy of a hydrogenic ion - Wikipedia.

[2] - Slater's rules - Wikipedia