Why do batteries drain so quickly if they are forced to draw more current than they normally do?
The capacity of a cell is measured toward the terminal cut off voltage. If e.g.a double current is drained from a cell, the external cell voltage is decrease by the doubled voltage drop on the cell internal resistance: $$U_\mathrm{e}=U_0 - R \cdot I$$
That caused reaching the terminal voltage sooner then in the half time.
It is additionally leading to the ohmic internal energy wasting $$P=R \cdot I^2$$
But this energy loss affects the useful energy output, not the capacity ( directly).
Additionally, higher than optimal current causes lowering of the immediate cell open voltage. Limited rate of diffusion of electroactive components and/or of electrode reactions are not able to maintain electrode equilibrium potentials.
In result, even the $U_0$ (non linearly) decreases with the higher current and the terminal $U_e$ is reached after providing smaller total charge.