Where there is a high-impedance supply, extra elements in parallel will cause the voltage to be reduced. The high-voltage pulses come from comparatively high impedance sources, so it doesn't take a lot of current to reduce the voltage.
The pulse specification of +100 V /-300 V is with no load. Any load, whether it's a TVS or a lamp will reduce the voltage.
In the example of a negative pulse, that can come from a relay coil when the supply is removed. The relay coil has significant inductance and so the current can't change instantly. The -300V is the worst case when there is nowhere for the current to go except to charge up some stray capacitances.
It's quite common to add a resistor in parallel with the relay coil to reduce the voltage. Here is an example data sheet:-
https://www.farnell.com/datasheets/1934206.pdf
On the standard 12 V relay, the coil resistance is 90 Ω, and there is an option of a resistor which is usually 560 Ω. If a relay coil like that is suddenly disconnected, the current in the coil will be 12/90 = 133 mA. That current will flow in the 560 Ω resistor, giving a voltage of around 75 V. When that happens the resistor has limited the voltage rise to 75 V. Without the resistor, the voltage rise would have been much larger. With a known limit of 75 V, it's quite easy to specify a suitable transistor to switch the coil current.
It is that type of voltage surge that the standards are there to protect against. A 1.2 W filament light will have a resistance of around 120 Ω, so if one of those were in parallel with the relay, it would limit the voltage surge to around 16 V, which isn't significant. Larger lamps would reduce the surge voltage further.
Resistors like that are often used in preference to diodes, because they can't be connected the wrong way round and because the current in the relay coil falls much faster with a resistor than with a diode. That means the that the contacts open faster, reducing arcing.