Maplin heated clothing
I've seen a few comments about buying the Maplin (battery powered) vest and converting it to run off the bike "because it's cheap", with the comment "9V is close to 12v so it can't be that hard" !!
No is the most likely answer. This jacket runs off two sets of 3 x AA batteries, so each pack is only 4.5V. Most likely the heated wire elements are set to run off each side independently so the overall wire resistance can be low and also get a reasonable battery life. So making a guess that this jacket generates say 10w from each battery pack (any more would pull too much current), the wire resistance that would be attached to get that power dissipation is about 2ohm. (P=Vsq/R). This would draw around 2.25A current (V=IR).
Upping everything to 12V will generate 70W !! Given bike voltage is more like 13V then the power would be even higher. Power is related to the square of voltage, not simply linear. If your fuse doesn't give out then it's goodbye jacket and you in flames !
Bike heated clothing also has to cope with a lot of wind chill factor so their typical power dissipation is around 45-50W, warm enough without burning. The electronics is designed around this, namely the wiring element, to dissipate that power from 12V to 14V needs to be a resistance of around 3ohms drawing a current of about 4A max.
I've made some assumptions but the message is it's not that simple to take lightweight garments like this and make them bike suitable.
If you google DIY heated clothing, you can find various instructions about how to make it. Maplin do a pulse width modulator motor speed controller that can be used to control the 12V bike supply and get a variable heat control. But the Teflon coated wire of resonable resistance is only available from RS components (a trade supplier). You will then be faced with working out how to get about 30-40 feet of cable inside a jacket !