If the mixture is being poured back into the jug with water, then how can the jug with water also contain ${\frac {V}{V+Q}}$. I would assume that when the diluted alcohol is poured into the jug with water, that water jug will contain ${\frac {V}{V+Q}}$ plus water.
You must make the distinction between the concentration and the volume. Let's take a look at a numerical example to see that the description is correct.
Suppose we start with $200$ mL of pure alcohol in one jug and $200$ mL of pure water in the other. ($V=200$ mL)
Now, we pour $50$ mL of water into the alcohol jug. This leaves $150$ mL of water in the water jug, and now we have $250$ mL of liquid in the alcohol jug. In particular, the liquid in the alcohol jug is a mix of $50$ mL of pure water with $200$ mL of pure alcohol. That is, the alcohol's concentration is now $$\frac{200}{200+50}=\frac{200}{250}=80\%$$ (and $20\%$ water) even though the same volume of alcohol is in the container as before.
Now, we mix it thoroughly, meaning that we are to assume that for any volume $V_0$ of fluid taken from the alcohol jug, it will be $80\%$ alcohol and $20\%$ water (as will the remaining volume in the jug). So, when we pour $50$ mL of mix back into the water jug, we are left with $200$ mL of the $80\%$ alcohol mix in the alcohol jug, and have poured a mix that is $40$ mL alcohol and $10$ mL water into the water jug. That amounts to a total of $160$ mL water and $40$ mL alcohol, so the fluid now in the water jug is $$\frac{160}{160+40}=\frac{160}{200}=80\%$$ water (and $20\%$ alcohol).