The shunt wire wrapped around the lead-in wires has a special insulating coating which has a high resistance. It isn`t a complete insulator, but is nonetheless enough that it becomes inconsequential when the main low voltage filament is there in paralell with it.
When the main filament burns out however, the voltage accross the lead-in wires rises to near that of the incoming supply, since at that moment almost no current flows through the rest of the bulbs in series. With this high voltage present, now the resistance of the coating of the shunt wire becomes more significant. Enough current flows to cause it to heat up and melt at the points where it touches the leads. Once melted through, the leads are essentially shorted together by the shunt wire, bypassing that bulb and lighting the rest again. This all happens extremely quickly when a bulb burns out so it appears almost as if the supply to the rest had never been interrupted.

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