The diagram explains everything. A normal diode does not conduct in the reverse direction so if you wire them as shown and the signal is between (-) and (+) they do nothing at all because the voltage is always 'backwards'. If the voltage goes below (-) which in most cases will be ground, the bottom diode becomes forward biased and it conducts, sinking available current to ground. Similarly, if the input voltage goes above (+), the top diode becomes forward biased and conducts current into the supply rail. Either way, it prevents the input pin of the IC going too positive or too negative and damaging it. The word 'transient' implies it is for a short period but in reality, the circuit works all the time, regardless of how long the over voltage lasts. Note two points though:
1. It relies on there not being enough current at the input to damage the diodes or to lift the (+) rail higher than normal.
2. Even a forward conducting diode drops a small voltage (typically 0.6V to 0.7V) so it will let the input voltage go to (-)-0.6 and (+)+0.6 before conducting which can still damage some ICs. For this reason, Schottky diodes are often used because the have a lower forward voltage, typically 0.3 to 0.4V.
Brian.