You could use a Polyimide etched-foil thin-film heater such as those made by Minco.

As far as refrigeration goes, if this was a serious high-volume application I'd be looking at something other than Peltier modules, probably completely out of the EE realm. Peltiers are just too inefficient and have too much thermal mass.
I can't tell you what the problem is with your apparatus since the photos can only give a limited amount of information.
However, it appears from the photo that you just have a voltage source (battery) in series with a resistor in series with an ammeter (you do have the multimeter set up to measure current and the leads connected to the proper inputs, correct?).
Now, this is certainly a valid connection and, if the above is a correct description, you should be able to measure different current values for different resistor values (however, if your multimeter and leads are set to measure voltage instead, you will only measure the battery voltage regardless of the resistor value).
However, with such a simple circuit, you can simply use Ohm's law to calculate the current.
Since an ammeter is effectively a short circuit, the current through the resistor is simply the battery voltage divided by the resistance.
In other words, if you simply connect the resistor directly across the battery and then measure the voltage across the parallel combination, the current is simply the measured voltage divided by the measured resistance.
Best Answer
Some nichrome wire run at low current could be made into an arbitrarily shaped element.
Conductive thread of some sort may be the best option if you can make it work.
Article on heated socks using conductive wool here.
Link to circuit used for heating thermochromic ink on fabric.
At this temperature a few resistors could easily be used too.
The Nichrome link, and resistor datasheets will give thermal data. Here is an example from a resistor datasheet:
This can be used to work out the voltage you would need to apply to dissipate the power for your desired temperature rise.
For example the CFR25 size would need to dissipate 0.125W to reach 25 degrees C above ambient. This is assuming the ambient is 20 degrees for a total of 45 degrees.
Say your supply voltage is 12V. For 0.125W we need 0.125/12 = ~0.010A. So the resistance needed would be 12V/0.010A = 1200 ohms.
For a ready made solution, a heater pad like this could be used. Various other sizes here.