Well, it may be such an old question now that the answer is moot, but as an electronic technician, let me address a couple of the points you're dealing with.
First, your charger is most likely putting out rectified, unregulated DC voltage. What that means is that it's taking the AC wall current, and either half-wave or full-wave rectifying it without any circuitry to smooth out the resulting pulsating DC to a clean, pure DC voltage. When you read this with a voltmeter, you're reading an average, not a peak reading, so the 8.2 volts seems about right. Think of the ocean waves, with peaks and valleys between them, and drawing a line that defines the average of the volume of water represented in those waves, and you have a basic idea of what the voltmeter is doing with a pulsed DC output. My guess is, if you read your charger output with an AC voltmeter, you'll read higher than 8.2 volts.
Most wall chargers rely on the batteries themselves to act as regulators/filter capacitors. This fits what you're seeing in the results. If the charger were only outputting 8.2 VDC, and the output were a smooth, clean DC voltage, then your battery could never charge above 8.2 VDC.
The rechargeable battery pack, if it's 8 NiCAD or NiMH batteries, is going to accept a nominal charge of 9.6 VDC (rechargeables typically read 1.2 VDC each, rather than the 1.5 VDC of comparable alkaline batteries). Both NiCAD and NiMH batteries have limited recharge cycles, and NiCAD's have a nasty trait of developing a "memory", where if you charge it, use it for an hour, and then recharge it, it will eventually reach a cycle of not lasting more than an hour in use, even if the rated output would have it lasting for several hours. NiMH are not prone to doing this, though it's still better with most rechargeable batteries to use them until they're starting to get weak, and then recharging.
Your momentary short may have damaged one or more of the battery cells in the battery pack, and that fits what you're describing much better than the charger itself being damaged. A bad cell may appear to take a full charge, but doesn't hold it at all. What happens is that a cell damaged in this way will take a full voltage charge, but not a full current charge.
Batteries deliver power, and electrical power is a product of two factors - voltage and current. Power, measured in watts, is derived by multiplying voltage by current. Ergo, a 1.2 volt battery with a current capacity of 1200 mAH (milliamp hours), should be able to deliver 1.4 watts of power for one hour. It can deliver higher wattage for a shorter time, or lower wattage for a longer time, based on the current draw. When you gang the batteries together, the wattage potential of each battery adds, so 8 cells capable of 1.4 watts for one hour would give you the potential for 11.5 watts for one hour. If you're only drawing 5 watts of power out, they should be able to deliver that for over 2 hours. If you're drawing 23 watts of power, they're going to be drained in a half-hour.
Rather than delivering its voltage at a 1200 mAH potential, a bad battery may only be able to deliver 1 or 2 mA. Your light requires far more current than the battery is capable of delivering, and it dies instantly. Once dead, it acts as a block to the power of the remaining battery cells, and your pack appears completely dead. If your battery pack can be opened, an outfit like Batteries Plus should be able to identify which cell is bad, and replace it. Whether this would be cheaper than buying a new pack directly from UK remains to be seen. If you have a Batteries Plus in your area, they should be able to test the battery pack itself and tell you if it is bad. From the sounds of what you described, though, my bet would be the battery pack.
