Why isn't ΔH = TΔS in ΔG = ΔH - TΔS?
However: $\Delta S_{sys} = \frac{q}{T}$, assuming the process is reversible (as usual).


No. $\Delta S_{env} = \frac{q}{T}$, assuming the process is reversible (as usual). So you get $\Delta S_{universe} = -\frac{\Delta G}{T}$ for a process where the heat exchange with the environment is reversible (and there is no non-PV work).

For more background, see https://chemistry.stackexchange.com/a/114323