Time position in STFT output
Assuming 0-based vectors, hop size $H$, and window size $M$, the frame number $N_f$ will contain the DFT of a window that starts on sample $n_1 = N_f \times H$ and ends on sample $n_2 = n_1 + M -1$.

That means that for each STFT frame, you shift the window by $H$ samples. Note that $H$ and $M$ not necessarily coincide.