How does this method of calculating the water equivalent of a calorimeter work?
So you're mixing hot and cold water in a calorimeter, and from the temperature change read from a time versus temperature curve, you're finding the heat capacity of the calorimeter. You're not understanding why the heat capacity is computed the way it is. 

First, identify all of your heat flows. You've got heat coming from the hot water, heat absorbed by the cold water, and heat absorbed by the calorimeter. If energy is conserved, $$q_{hot} + q_{cold} + q_{calorimeter} = 0$$

One thing that's confusing you is that last term, $q_{calorimeter}$. It isn't zero. It's very small, since your calorimeter is Styrofoam, but it will be measureable. Measuring it is a central goal of this experiment!

Now, you have a time versus temperature curve for the contents of the calorimeter. You're extrapolating the temperature plateaus before and after mixing to the time of mixing to figure out the initial and final temperatures. Since you had the cold water sitting in the calorimeter initially, and you dumped the hot water into it, the final temperature minus the initial temperature gives you the temperature change for the cold water. It also gives the approximate temperature change for the calorimeter if you can assume that the temperature of the calorimeter is the same as the temperature of its contents.

The heat capacity of the calorimeter is $$C = \frac{q_{calorimeter}}{\Delta T_{calorimeter}}$$
You know $\Delta T_{calorimeter}$ from your time vs temperature graph. You know $q_{calorimeter}$ from the energy conservation equation above; $q_{calorimeter} = -q_{hot}-q_{cold}$. Therefore you know $C$. You should know how to get $q_{hot}$ and $q_{cold}$ from the masses and initial and final temperatures of the hot and cold water (and the specific heat of water). Can you take it from there?