Understanding a phase bump filter
Yes, it is just transfer function math. If you have a transfer function, you can compute its phase as well as it's gain at different frequencies. For a capacitor:

$$T(f)=I_o/V_i = \frac{1}{sC} = \frac{1}{j\omega*C}$$

Where $$j$$ is your imaginary number. It means a phase shift of 90. It's reciprocal is also -90. Suprise, surpise, in a cap, current lags voltage by ninety degrees. 

For a more complex transfer function, we can compute phase as follows. 

$$T(F) = V_o/V_i = \frac{1}{1+sC}= \frac{1}{1+j\omega*C}$$

Choose a capacitor value of C = 1. In this case, we see that our phase will be frequnecy dependant. Give 6.28 Hz, for convienience, this would be 1 rad/s and so:

$$\omega = 1 $$

$$T(F) = V_o/V_i = \frac{1}{1+sC}= \frac{1}{1+j\omega*C} = \frac{1}{1+j}$$

$$\angle T(f) = \angle Numberator - \angle Denominator$$

$$\angle T(f) = 0 - tan(1/1) = 0-45 = -45 degrees$$

Thus this low pass filter will also introduce a lag of 45 degrees at 6.28 Hz!

After I began typing I realized that my explanation was a little math heavy. If you're unfamiliar with imaginary numbers, this is a bit of a pain. But you did say you'd worked out a transfer function. Makes sense to me though! If you'd like a specific example, feel free to ask for a walk through. Also, if it's unclear, I can always add more detail. Good luck!

From a less mathematical view, you're using a high pass and a low pass together. The high pass will lag at low frequencies, and the low pass will add lag at high frequencies. In your pass band, it won't add as much lag to this frequency, thus saving you 15 degrees. 

For suggested reading, I would say calculating transfer functions, not just measuring them.