Would you mind showing this? A visual would be great
The peak voltage on the transistor will be increased by a factor of R100/[relay resistance] so you must make sure that the peak voltage is still less than the transistor maximum Vce.
The resistance of the G5LE relay is 39 Ohms. The current at 5 V is 128 mA. When the transistor turns off, that current will flow through R100. If R100 is 220 Ohms, that will give a voltage of 28.2 V across R100. The voltage between the collector and emitter of the transistor is that 28.2 V plus the supply voltage plus the diode voltage, so about 34 V. That is OK for a transistor rated at 40 V or more.
The resistance in the freewheel circuit is 39 + 220 Ohms, which is 6 - 7 times larger than it would be if the diode were directly in parallel with the relay coil, so the current in the coil will reduce about 6 - 7 times faster.
The LED current is limited by both R100 and R101. If the LED voltage is 2 V, R101 is 0 Ohms and R100 is 220 Ohms, the LED current is 13 mA. R101 can be increased to reduce the LED current if required.
The connection with R100 and the LED in parallel with the diode instead of in parallel with the relay coil prevents the reverse voltage spike of 28.2 V being applied to the LED.
In your application it may not matter if you turn off the relay coil slowly. It will still appear instant to someone watching. The curves in the data sheet show that the life of the relay is reduced quite a lot as the load current goes up, and inductive load reduce the life.
Many automotive relays have a resistor built in, in parallel with the coil, and no diode. The transistors that drive the relays have to be rated to 100 V or so, if there is no external diode, but they probably would be anyhow. Having no diode means that the coil polarity doesn't matter.