The coil can work from any polarity.

And the coil must work down to 3.75V (but may possibly work at an even lower voltage).

In your case, you burn 1.5V over your 120 ohm resistor. That is 12.5mA.

You have 3.5V for your LED + resistor. Where id you get a 260 ohm resistor by the way. Is it a 270 ohm resistor that you measured with your multimeter? Anyway, the diode has about 2V Vf, so the voltage over the 260 ohm is about 1.5V. That means that the current through this resistor is about 5.8mA.

The remaining current - (12.5-5.8) = 6.7mA will then go into the relay. The datasheet says that the relay coil has a resistance of 500 ohm, and 3.5V and 500 ohm should be about 7mA which seems to agree with the expected current.

The next thing is that we don't really know exactly how any internal diode is connected. The text in the datasheet - saying there is a zener diode in series with a normal diode - sounds a bit strange, and may be the result of a copy-paste accident.

Anyway. If the polarity is correct, there will be no current through any internal diode. If the polarity is wrong, and there is just a normal diode inside the relay, then that diode will start to conduct somewhere between 0.3 and 0.7V depending on if it is a schottky diode or a normal diode, and depending on the size of the diode. It is possible to burn this diode by applying the wrong polarity, and then continue to use the relay even after the diode is dead. But with the dead diode, there will be no free wheeling protection. But for a little reed relay, it is possible that the diode may support more current than the wiring inside the relay, so applying the reversed polarity may burn the connection to the coil instead of burning the diode.

If you have the diode in series with a zener diode, then you can increase the voltage where the diode will start to conduct, which means that you will not be able to measure any extra current through the diode even if using the wrong polarity - until you get above the "knee" of the diode combo.