Why does my digital multimeter report small voltage between a battery and unattached metal and my fingers?
On some ranges, a digital voltmeter (DVM) may have very high input impedance. A small static charge builds up on any non-grounded conductor. Atmospheric electricity is always present. The potential averages about 100 V/m height difference, so there is might be a 200 V difference between a 2 m tall person's head and toes. Of course, the wet, salty body is a fairly good conductor, so it effectively short-circuits most of the charge to ground... unless, perhaps, you're wearing rubber-soled shoes. So the DVM could have been measuring charge buildup.
A DVM also may be subject to self-rectification of AC pickup. Even though the scale is set to DC voltage measurement, RF pickup from radio stations and 50 or 60 Hz pickup from the power mains can still show as an erroneous DC reading.
As an experiment, I just set an inexpensive (US$6) DVM on the 200 mV scale, separated the leads at arms length, but with no contact to the probe tips, and walked around the room. Motion on the carpeted floor generated a large static charge, that bled off in a few seconds. Placing the leads near a WiFi router picked up a small signal, i.e., accidentally rectified RF.
You can make an even higher impedance voltmeter, an electroscope, with household materials. BTW, a spark jumps an air-gap at, very roughly, 1,000 V/mm, so a 1 cm spark made by walking across the floor and reaching for a grounded object might be 10,000 V. Shocking!
But you ask could you power something. It depends on the current (I), i.e., amount of electricity (compare with amount of water, whether a drop or a bucket full), as well as the voltage (compare with water pressure). Wet human skin, in series with the body, may have a resistance as low as ~500 ohms, and dry skin as high as 100,000 ohms. Resistance impedes, or partially blocks electricity, so you would not have much current, measured in amperes (A), milliamperes (mA, 1/1,000 A) and microamperes (µA, 1/1,000,000 A). Current = voltage / resistance (I=V/R), so at 12 V DC with 100,000 ohms, that would yield a current of 0.00012 A (0.12 mA), which could light a LED (light emitting diode) dimly. It would be safe to hold one lead of the LED, touch a finger to one side of the battery and the other LED to the other side of the battery. If it doesn't light (dimly, so this won't be visible in daylight or a brightly-lit garage), reverse the leads of the LED.
There would not be enough current to power an incandescent lamp, motor, or almost any other device than a lone LED, or perhaps a LCD clock, though.