Why is such amplifier configuration chosen for HB100 microwave sensor?
Figure 1. The various filter componenents.


The 4.7u and 330k at (1) form a high-pass filter. Cut off is at \$ \frac {1}{2 \pi RC} = \frac {1}{2 \pi \cdot 4.7 \mu \cdot 330k} = 0.1 \ \mathrm {Hz} \$.




Figure 2. An active high-pass filter. Source: Electronics Tutorials.

The RC networks (2-3) and (4-5) form active band-pass filters.


The low frequency roll-off is at \$ \frac {1}{2 \pi \cdot 4.7\mu \cdot 1M} = 0.034 \ \mathrm{Hz} \$.
The high-frequency roll-off is \$ \frac {1}{2 \pi \cdot 2n2 \cdot 10k} = 723 \ \mathrm {Hz} \$ for the 10k and you can work out the 8k2 version.



  One friend suggested that a differentiator forms a band-pass filter and an integrator + differentiator combination cancels each other out, leaving you with a linear amplifier with a band-pass filter.


The integrator and differentiator would only cancel out if they had the same cut-off frequencies. They don't so you get a band-pass filter.


  If that's that case, why would we want that?


The datasheet explains on page 3: The magnitude of
the Doppler Shift is proportional to reflection of transmitted energy and is in the range of microvolts
(µV). A high gain low frequency amplifier is usually connected to the IF terminal in order to amplify
the Doppler shift to a processable level (see Annex 1). Frequency of Doppler shift is proportional to
velocity of motion. Typical human walking generates Doppler shift below 100 Hz.