It looks like both filter sections are designed with the -3dB point at the same frequency, or very close together, so this filter is doing what it should.
In the crossover region, both sections contribute to the output, so it is higher than either alone. The slight peax at the crossover frequency would be 3dB if both signals were in phase(so they added coherently), so presumably they aren't. EDIT : apparently the small separation between -3dB points, rather than phase, accounts for this peak being less than 3dB.
For a classic design without that bulge, read up on the Linkwitz-Riley crossover, commonly used in loudspeakers where you want HPF and LPF outputs to sum to unity.
I don't know what you were expecting but if you wanted a notch you'd have to separate the -3dB frequencies, then the depth of notch will depend how far apart they are, and it won't be a deep notch.
If you wanted a deep notch, one approach is the Twin-T filter which can be made as narrow as you want.
Or start by specifying the frequency, notch width (at -3dB), and notch depth you want, and research filter design techniques to meet that specification.
I think what you are asking is if the two circuits in your diagram are equivalent.
If so, the answer is yes. The Thevanin equivalent of the bottom network attached to the non-inverting input of the amplifier is exactly the top network. So they are identical in operation, at least ideally.
Best Answer
I think put the capacitor as a degeneration as in the following circuit:
simulate this circuit – Schematic created using CircuitLab
Here, A and B are the input nodes of your sensing element as you show in the figure. \$vo_+\$ and \$vo_-\$ are the differential outputs. For frequencies below (\$\frac{g_m}{C}\$), the gain of the circuit is proportional to the capacitance. $$A_v = \frac{R}{\frac{1}{\omega 2C}} = 2\omega RC$$
Based on this gain capacitance can be calculated.
EDIT
The above circuit is not an opAmp circuit. But as LvW said, you can use differentiator with OpAmp to get the gain proportional to capacitor: $$A_v = \omega CR$$
simulate this circuit