The voltage across the ends of a battery is estimated by $$ V = E - I r$$ where $E$ is the volt rating, $I$ is the current and $r$ is the internal resistance of the battery.
So with two batteries $E_1$ and $E_2$ connected in parallel to some load $R$ they must share a common voltage
$$ V = E_1 - I_1 r_1 = E_2 - I_2 r_2 $$
and the total current through the load is the sum of the battery currents
$$ I = \frac{V}{R} = I_1 + I_2 $$
with solution
$$\begin{aligned} I_1 & = \frac{E_1 (R+r_2)-E_2 R}{R(r_1+r_2)+r_1 r_2} \approx \frac{E_1 -E_2}{r_1+r_2} \\ I_2 & = \frac{E_2 (R+r_1)-E_1 R}{R(r_1+r_2)+r_1 r_2} \approx \frac{E_2 -E_1}{r_1+r_2}\\ V & = \frac{R (E_1 r_2 + E_2 r_2)}{R(r_1+r_2)+r_1 r_2} \approx \frac{E_1 r_2 + E_2 r_1}{r_1+r_2} \end{aligned} $$
So $I_1$ and $I_2$ are equal and opposite to each other. If $E_1-E_2$ is postive (first battery has more voltage) then $I_1$ is positve (supplying power to system) and $I_2$ negative draining power.