Originally Posted By: Johnny2Bad
From your bottom photo it looks like they are connected in parallel, with a 0V / +6V supply. This is done to increase the current capacity of the battery powered device. That the lid operates but not the motor seems to support this ... the device requires more current than a single bank can provide to fully operate.
However you should confirm this with a DMM ... be sure the + lead of bank 1 is connected to the + lead of bank 2.
IF they are connected in parallel, now your problem is to discover (calculate) the current requirements, and then seek out a +6V wall wart with the appropriate current rating (in mA [miliamps] or A [amps]). If you have trouble sourcing something exactly correct, it's OK to use a replacement supply rated higher than you need but not lower.
Estimate 2000 mAH per 1.5V cell, so 8AH per 6V bank, total 16AH @ 6V.
With a 6V supply that translates to about 2.67A capacity, so that's the size of wall wart you should seek. It will probably work with a smaller capacity wart but you don't know how much smaller (as it eats batteries) so don't cheat too much there.
You are also going to have to do some soldering, one way or another, so hopefully your skills are up to the task and you have a decent iron. Ordinary 60:40 or Eutetic 63/37 tin/lead solder is fine. Lead Free requires higher temperatures and is easier to damage components with, but if you expect to be exporting your device to Holland or Germany, by all means use EU-legal lead free.
Thanks for your valuable input and advice