In my first answer I have described how you can find the exact solution for the 2 zeros and the 2 pole frequencies (which are identical to the wanted break frequencies).
However, here is a good approach which should be sufficient for the shown circuit.
In principle, I follow the way as outlined already by Dave Tweed´s answer: Simplification of the circuit.
In the present case, you can create three different (simplified) circuits of first order only which easily can be analysed.
1.) For the first rising region of the transfer function the high pass part with C1 is responsible (C2 causes the falling part and can be neglected). Furthermore, for very low frequencies (including DC) the gain Ao=1+R3/R2 is assumed to be not much larger than unity which is the possible minimum.
Hence, for acceptable filtering it is assumed that R2>>R3.
As an equivalent diagram for the lower frequency range (without C2 and R2) we arrive at a circuit with only the three components R1, R3 and C1. It is a simple task to find the relevant time constants (invers to the corresponding break frequencies):
Using your indices, we thus find T2=(R1+R3)C1 and T1=R1C1.
2.) Above the frequency f1 the capacitor C1 is not effective any more (and the capacitor C2 is assumed to be not yet effective). Hence, we have a simple non-inverting amplifier with the gain (maximum of the transfer function) Amax= 1+R3/Rp with Rp=R1||R2.
3.) For rising frequencies, the low pass part with capacitor C2 becomes effective (C1 is considered as a short). Hence, the feedback path consists of R3||C2 and Rp only.
The time constant T3 (pole frequency) can be derived as T3=R3C2 and the last break frequency (zero) is determined by T4=R3C2/(1+R3/Rp).
Finally, it is to be noted, that all results are in agreement with the values given in the scetched BODE diagram. This can be verified using the well-known relations for a 20dB gain slope (as used in the graph with G1/G2=f1/f2).
Final remark: Thus, it can be concluded that the information contained in the scetched BODE diagram (break frequencies) also are only approximations.
The only issue with the adapters could be a mechanical one. They are usually pretty short, so if you don't have good access to the connector it might be difficult to screw it in. Also, you could run in to an issue where the antenna is screwed on to the adapter and then the adapter gets loose, but you can't unscrew the antenna because the adapter rotates and you can't access the adapter to either tighten it or loosen it. So if you can access the connector easily even with the antenna installed, the adapter is probably a good option since you already bought the antennas. Otherwise, it might be a better idea to buy the correct antennas.
Best Answer
The filter is not directional. Passive filters are inherently reversible, though not always cascadable.
The different connectors on either end do not imply any particular direction, they are simply to allow the filter to be dropped in at the end of a cable as a bolt-on, often as part of troubleshooting.
Male is the connector with the pin in the middle.
If you are only using it to try and listen in to some transponder signals while you are on the ground, then this will work fine. ADS-B usually has quite a generous link budget owing to the massive Tx power of the transmitters.