P-Channel Mosfet overvoltage protection using a TVS

By placing a zener diode (or similar) between the drain and gate, the gate voltage will be forced to rise if the drain exceeds the breakdown voltage of the diode. You are correct in thinking that this will turn the MOSFET on. This is how that protection method works. To see why you have to understand MOSFET avalanche. The body diode of the MOSFET will break down at some voltage like a zener diode (or avalanche diode to be strictly correct). A power MOSFET is made up of many cells in parallel with each other. Depending of the processing of the transistor, these cells may or may not share the avalanche energy well. Also the place within the cell where the avalanche energy is dissipated is not quite the same as where normal conduction dissipation occurs. For this reason some MOSFETs have a limited amount of energy that they can safely absorb in a single pulse of avalanche. This energy may be much less then they could take in a pulse of normal conduction. It may be zero. On the other hand, some MOSFETs have avalanche energy limited only by thermal considerations, in other words they can take any amount of avalanche energy as long as the die does not overheat.

By turning the MOSFET on before the drain-source voltage exceeds the avalanche point, the energy is dissipated by the normal conduction mechanism. This allows a MOSFET which cannot take high avalanche energy to safely dissipate a transient, for example from and unclamped inductive load. You still have to ensure that in your design the MOSFET cannot be exposed to a transient that is too large from a thermal point of view, the energy will heat the die up just the same and you must not exceed Tj max.

You have to be careful adding diodes to the gate terminal. In some circumstances diodes connected directly to the gate can result in very high frequency oscillation of the MOSFET. This results from parasitic inductance and capacitance in the circuit causing a feedback path at VHF. A diode can rectify this and pump up the DC level on the gate resulting in a stable oscillation maintaining the device in a partially on state when trying to turn it off. This is mitigated by adding a suitable gate resistor.

If your circuit topology ensures that the MOSFETs will never see excessive voltage, or if they can withstand any transients by avalanche then you may well not need to provide any protection components. This is a good solution because there can be more problems introduced by the protection devices (too big a subject to cover here). Decoupling the supply rails near to the output devices in power stages helps. Ensuring that devices in bridge configurations cannot cross-conduct is also very important.

The second circuit may blow-off either the BJT or the zener (or both), if 12V_IN actually is larger than 15V...

You must add a resistor as shown here:

To answer your question:

1. The most reliable way is the zener, provided that you put the resistor as indicated.
2. 1k not only is enough, but might be unnecessarily low, if you don't want high speed switching. Much larger values (>>10k) can be used if you are going to use this circuit to power on/off some devices in your system.
3. Voltage does not flow. There will be damage if you apply a high voltage (i.e. higher than the voltage supply of your MCU) to a GPIO pin or if you exceed any voltage rating of your MOSFET.
4. If \$V_{GS,max}\$ is 20V, then you should make sure that this voltage is not exceeded. You can't use a 20V zener (Zeners, like all components, have some tolerances). Your 15-V zener is ok, but also a 10-V zener would be ok.