What a beautiful question! So simply posed, but look at the diversity of long-winded answers. Let me add my own....
Why is the amount of work required to accelerate a body from 10m/s to 20m/s three times the work needed to accelerate a body from 0m/s to 10m/s?
I think the first thing to point out is that there's no "physics" involved here. It's just the simple mathematical identity
∆(v2/2) = ∫adx
dressed up with the words "work" and "kinetic energy."
Explicitly, d/dt(v2) = 2av, so ∆(v2/2) = ∫avdt = ∫adx. Multiply by m and you get ∆KE = W. (Or if you don't like calculus, you can show the same thing for the case of constant acceleration, with x=x_0+v_0∆t + a∆t2/2 and some algebra.)
So "work" and "change in kinetic energy" are literally just two names for the same exact thing. Your intuitions "harder to accelerate" and "do more work" are in some sense just a distraction.
If you strip away the work-energy concepts, then your question can be recast as a comparison of the identity
∆(v2/2) = a∆x
in two different reference frames, one in which the velocity goes from 10-20 m/s, and another in which the velocity goes from 0-10 m/s. That would be a good exercise.
Finally, even at a fundamental level, you can regard the work/energy concepts as mere bookkeeping devices that allow you to write velocities as a function of position, instead of time.
That's a profoundly useful thing to do, because if you know the force as a function of position (you always do!), then you also know velocity as a function of position. Instead of talking about acceleration over time due to forces, you talk about "potentials" that create changes in "kinetic energy" due to position. Just a change of variable! But in particular, all the fundamental forces depend only on the relative positions of things (no explicit time dependence), which implies that microscopically, the velocities within a system depend only on its configuration.
Energy, baby! But in a sense, it's all just the same trick, ∆(v2/2) = ∫adx