Resistors are useful before and after the TVS to serve different purposes. The cap may be placed either in parallel with the TVS or directly at the processor pin; the latter will provide somewhat more protection, but will also cause the processor to respond more slowly to changes in the input.
If the input to the device were connected to a charged-up capacitor (e.g. 100V)
and there were no resistor anyplace, the TVS might quickly clamp to 6V, but
the processor's internal protection diode would have a very large amount of
current forced through it with a one-volt drop. The vast majority of the
energy from the capacitor would be dissipated in the TVS, but the processor
would still absorb a damaging amount. Further, almost all of the energy
would need to be handled by the TVS.
Adding a resistor between the outside world and the TVS would reduce the
current, but since the resistor would have almost 100 volts across it, it
would pass a significant amount of current, and that current would end up
flowing through the chip's protection diode. As above, the TVS would help,
but leave a large amount of energy for the chip to handle. In this scenario
most of the energy would be dissipated by the resistor rather than the TVS,
so the TVS would be stressed less severely.
Placing a resistor between the TVS and the chip, but not between the TVS
and the outside world, would protect the chip provided the TVS was able to
clamp the voltage effectively, because the resistor itself would only have
a few volts across it. The TVS, however, would be taked with dissipating
almost all of the energy from the capacitor.
Placing a resistor on both sides of the TVS would provide by far the best
protection. Most of the energy would be dissipated in the first resistor,
making it much easier for the TVS to absorb the rest, while the second
resistor would limit the peak current fed into the CPU.
simulate this circuit – Schematic created using CircuitLab
The circuit above can be simulated, with the relays opened and closed to show different combinations of resistors being present and absent. Use the "Simulation" button and "Time domain" tab, and "run time domain simulation". The top trace shows the current in the suppression diode and the "chip" [simulated at the right by a diode and resistor to VDD]. The bottom trace shows the current through the right-hand protection resistor; that will be zero when the relay is shorting out the resistor, but it shows the current in milliamps rather than amps. Adding the first resistor greatly cuts the total amount of current absorbed by the suppression diode and chip, but with just the first resistor the chip still has a rather high peak current. Adding just the second resistor would protect the chip pretty well, but leave the suppression diode absorbing a lot of current.