The datasheet says that pin 2 is +.

The polarity is important. When you supply your voltage to drive the relay, the diode should not short-circuit the coil. The diode is expected to short-circuit for currents in the reverse direction. When you disconnect the voltage to the relay coil, the inductance in the coil will try to continue to drive a current of the same strength through the coil. But to drive a current through a disconnected circuit (which has an infinite impedance), it will take a very, very high voltage.

If you short-circuit a small battery over a coil, and then disconnect the battery, you will see a spark when you disconnect. The battery has charged the coil with magnetic energy, and when you disconnect the current, this magnetic energy becomes a high-voltage discharge.

When you do not disconnect in a mechanical way, but use a transistor or buffer chip, that magnetic charge will force a current in the wrong direction through that transistor or buffer. Since the transistor or buffer will not lead, the voltage will climb and the energy will kill the transitor/buffer.

With a free-wheel diode, the coil will - instead of killing your buffer - instead be able to drive the current through the diode and back to the other side of the coil. The diode has a very low resistance, so the voltage will never climb higher than the Vf of the diode and you will never get that high-voltage destruction of your drive transistor/buffer.

Whenever you need to drive an inductive load, you have to make sure that the magnetic energy has somewhere to go. For bigger loads, that disconnect spark can be strong enough to kill.