Dealing with mains AC.
Some transformers are supplied with wire tails rather than tags.
One of those will allow mains connection with a minimum of danger. Solder the two lives mains lead wires to the two primary wires and use heat shrink sleeving over the connections. This can all be done with the mains never having been connected. When you are finished there is minimal chance of electric shock.
This does NOT include a fuse, which would be "good" [tm] to have. You can buy inline fuse holders which also have wire tails. You can do as above with Mains-fuse, fuse-transformer and transformer-mains joints, all soldered and all with heat shrink insulation.
DO NOT just twist wires together.
DO NOT use wiring twist on "nuts" which are solderless.
The latter can be very useful but are a very very very bad safety start when you are not used to mains.
Clamp mains lead with a cable entry clamp or several cable ties through several pairs of holes or similar so that there is no way for external mechanical stress
to be transferred to internal connections.
You can buy AC output plug pack transformers designed `for low voltage lighting use with ratings in the 1A to 2A range and voltages of typically 12 VAC to 24 VAC. This gives you a low voltage source of AC without having to deal with mains connections.
Note that 12 VAC has a peak value of about 17 Vdc. Easily enough for a low current opamp supply and enough for a DC supply at the DC = rated AC voltage as long as a sensibly low dropout regulator is used.
If you now use an eg transformer with a number of low voltage windings you can generate several windings. eg if you had a transformer with 2 x 24 VAC centre-tapped windings you can connect 12VAC to a 12V half winding and get 12-12 from the other centre tapped winding.
With a little Heath Robinson approach you can connect eg 24VAC across a centre tapped 24VAC centre tapped winding and then use the 24AC ct as 12-12.
The above transformers may also have a mains winding. Insulate it before starting and ignore.
You can connect an AC voltage to a winding intended for equal or greater voltage. eg 12 VAC into a 12VAC or 15 VAC or 20 VAC winding. If the target winding is too much greater than the input voltage the magnetisation current will be too low and the core will not be well used. Often not a problem.
You can ry connecting eg 15VAC to a 12VAC winding but the increased magnetisation current will drive the core towards saturation and even 15->12 is probably rather too much. If you try it and it starts to get more mildly warm you can probably disconnect without permanent damage. Probably.
LiPo batteries require complex charging patterns to charge the battery quickly and properly. The IC used in that package goes through a preconditioning, fast charge, and then constant voltage mode before it considers charging complete.
Drawing power from the LiPo while charging will interfere with this charging mechanism. It may work, but it will abuse the LiPo and probably affect its life. It may overheat.
Normally the charging circuit will power the system directly while also charging the battery. This can be implemented connecting Vcharger -> diode D1 -> Vin, and Vbatt -> diode D2 -> Vin. When charging Vcharger > Vbatt, so D2 will be reversed biased and the battery effectively disconnected. When there is no charger, D1 prevents the battery from energizing the charging circuit which is obviously not a good idea. If you are concerned about the voltage loss due to the diodes, the concept can be extended such that the diodes control the gates of power MOSFETs instead of directly passing the current.
Best Answer
It probably is feasible.
But: you need to realize that even for electricity, conservation of energy applies! So if you want to produce 10,000 Ws (wattseconds) = 10 kJ of heat, you need to have that much energy in your battery.
So, in your place, I'd start by calculating what amount of heat you want to produce. That's really easy. Calculate the amount of material (ie. flesh) you want to heat up, multiply that by the specific heat of the material (use the specific heat of water, flesh is mostly that stuff), and then you know how many Joule you need. One Joule is 1 W · 1 s.
Then divide that by the time (in seconds) you want to allow the heating to heat up that much material, and you've calculated the power (in Watt) necessary to heat up that amount of material that fast.
Now, I don't believe you know nothing. You've probably heard of 1 W = 1 A · 1 V. So if you have a 12 V battery, in order to produce 1 W, needs to make \$\frac1{12}\text{ A}\$ current flow. If you need 120 W, you'll need 10 A. It's that simple.
Now that you know how much current you need to spend, you can calculate two things:
This doesn't address things like maximum power impedance matching etc, but it does in theory give you a good idea of what your system needs to look like.