Physics check:

Two general components of that, in addition to m*g*h. One is accelerating the mass. The easiest way to visualize that is the difference in kinetic energy (1/2m v squared, for the two v's). That's how much it took to accelerate to the higher speed. The second part is that the drag is greater at higher speeds - approximately proportional to speed for rolling and drive train losses and the like, but more importantly proportional to the square of the speed for aero drag. Energy (that portion of it) is that drag force times the distance.

But the extra energy he needed to raise his mass up the hill was just m*g*h. It doesn't matter, for that energy, how fast you do it. The confusion here is that Armstrong is more concerned about power than energy, and power is how much energy is used in a given time. So twice as fast uphill takes twice the power, for just lifting the weight. And approaching eight times the power to overcome drag force from the air. If he's going fast enough, and I assume that Armstrong and Ullrich probably were.

But back to OP, the whole reason I brought this up is asking about hills and calorie burn. Bottom line, just add it in for the climbs, the formula I gave earlier, and that produces the (extra for the hill) calories burned. As opposed to flatter rides.