Why does this optocoupler circuit work without getting destroyed?
You may misunderstood reverse current, see http://www.renesas.eu/products/opto/technology/standard_p/index.jsp

The LED is a diode, so it is not intended to conduct in the reverse direction. However if you still force enough high reverse voltage to its pins, this very little reverse current does flow.

Scope (with proper insulation transformer) the voltage on the LED. LED - just like any other diode - does have a reverse breakdown voltage. This is the Vr in the datasheet. In reverse breakdown you can imagine the LED as a Zener, so once more than 4V applied in the reverse direction, current will flow.

Refer to this picture: http://reviseomatic.org/help/e-diodes/Led-graph.gif
You can read more at wiki: https://en.wikipedia.org/wiki/LED_circuit 

If you drive the led in reverse, performance of the optocoupler degrades over time, see http://www.renesas.eu/products/opto/technology/standard_p/index.jsp Vr.

Therefore it is a good idea to add a standard diode either in series (so no reverse current can flow), or to the led of the optocoupler in the reverse direction (so it shunts the reverse voltage).

Moreover, as this is a zero-crossing circuit, you can consider using a rectifier bridge then connect the led to the output of the rectifier. This results very clean zero crossing spikes in both halfwave.