If an alkene is stable, its in a "low energy state", so when it becomes more stable it will be in a "lower energy state". So, given that, if something goes from a high energy state to a low energy state it will have a higher heat of hydrogenation because its energy released was large as it went from high to low. And, if something was in a low energy state and goes into a lower energy state, the heat of hydrogenation will be smaller because it went from low to lower.
I believe heat of hydrogenation is an enthalpy value, so think of it in terms of enthalpy and I think it will make more sense.
ΔH = ΔE + PΔV - if we assume that there is no change in volume.
ΔH = ΔE = E_final - E_initial
So, when something is unstable it has a "high internal energy" and when it is stable it has a "low energy". So, the former case I presented in the first paragraph is indicative of a large, negative ΔE (Efinal << Einitial), and the latter case is indicative of a small, negative ΔE (Efinal < Einitial, but not "much"), so their heats of hydrogenation are large, and small, respectively.
Hope that helps.