At low temperature, the rotation around the N-C amide bond of dimethylformamide is slow, so the two methyl groups are in different chemical environments. The H NMR thus shows a peak for each of the methyl groups.
As the T is increased, we see that the peaks begin to coalesce, until eventually they coalesce into a single average peak. The basic explanation is that, since the methyl groups are exchanging their chemical environment at a rate faster than the NMR time scale, they are considered to be within the same average chemical environment.
I am not very comfortable with this explanation, because it could also be argued that 50% of the time the methyl will be emitting signal A, and 50% of the time it will be emitting signal B - and the FID signal should still contain two signals. Alternatively, if the rotation is continuous, then I could expect a very broad peak extending from A to B.
My guess is that there is something about the structure of the signal and the Fourrier transform that I do not understand. Could someone help me out understanding why this coalescence occurs?