I like answering these not because I know very much but because I usually learn something by thinking about it.
Normally you will have 4 beats to a measure where one beat is a quarter note. Normal
is why it is also referred to as C
or "common" time a.k.a. 4/4(hope I'm not making that up). Four quarter notes per measure. I like to look at it as the bottom note in the time signature is the denominator in a fraction. 4/4 = 4 1/4 notes or 4 * 1/4 = 4/4. For 12/8 you have 12 1/8 notes or 12 * 1/8 = 12/8. You literally have 12 8th notes per measure. You can always simplify to 6/4 or 3/2 (3 half notes). But it's always easiest to count your beats in either quarters or 8ths (it is for me anyways).
For actually playing that 12/8 time signature you will want to remember that the rhythm can be felt in threes (3/4, 6/4, 9/8, 12/8). In other words you can break up the measure into groups of three rather than 2 or 4. Seeing it as 6/4 makes this easier because it's two groups of 3 quarter notes per measure. It is the double of 3/4 time. To simplify just see it as a twice as long measure in 3/4 time.
Does this make any sense? Try wikipedia: http://en.wikipedia.org/wiki/Time_signature
According to wiki, 12/8 should be considered a compound time signature:
In compound meter, subdivisions of the main beat (the upper number) split into three, not two, equal parts, so that a dotted note (half again longer than a regular note) becomes the beat unit. Compound time signatures are named as if they were simple time signatures, in which the one-third part of the beat unit is the beat, so the top number is commonly 6, 9 or 12 (multiples of 3). The lower number is most commonly an 8 (an eighth-note): as in 9/8 or 12/8.
By dotted note they mean that you could see 12/8 as 4 dotted quarter notes (each dotted quarter being equal to 3 8th notes, 4*3=12). But who wants to count like that, seems like a lot of work. I would just look at it as [1,2,3, 1,2,3] or [1 and 2 and 3 and, 1 and 2 and 3 and] if doing 8th notes or however you count.
Not sure of my accuracy here but this is how I would approach it.