The basic surprise here is that energy is always conserved when waves superimpose. This seems like it shouldn't be true, since superposition causes amplitudes to add, whereas energy is proportional to the square of the amplitude. This nonlinear relationship makes it seem like energy should not be additive, and therefore can't be conserved.
This is a generic issue for waves, not just light waves, so let's consider it first in the case of waves on a string.
For example, if you take a light string and a heavy string and join them together end to end, waves that hit the boundary will be partially reflected and partially transmitted. One thing we can predict immediately is that energy is conserved, since the interaction between neighboring bits of the strings are governed by the ordinary laws of mechanics. Since energy is conserved at all points in space and time during the evolution of the waves, it must be conserved over all.
Another helpful example is a single uniform string, with oppositely propagating sine waves of the same wavelength encountering one another and superposing. There are moments when the waves cancel and the string is flat, but the motion recovers after such a moment -- was energy destroyed and then created again? No, because the flatness of the string only implied zero potential energy. The string had kinetic energy at the moment of flatness. If each wave separately had 1 unit of KE and 1 unit of PE, for a total of 4 units, then at the moment of flatness, we have 4 units of KE (due to doubled velocity) and 0 units of PE.
So in general, the idea is that conservation of energy has to be proved from the equations of motion, which are a differential equation holding at every point in space and time. For light, these equations of motion are Maxwell's equations, and one can indeed prove conservation of energy from them.
It's a nice exercise to verify conservation of energy for the case of oppositely propagating light waves, as in the second example with the rope above.
I can set up a thin film such that the reflectance is zero (R = 0) and transmittance is one (T = 1), don’t I need energy in the waves to cause destructive interference to begin with?
(I corrected T = 0 to T = 1 in this quote.)
Here we can analyze the reflected wave as a superposition of waves reflected from the front and back surfaces. Amplitudes add, not energies. Therefore these two waves, each of which individually would have had, say, 1 unit of energy, together do not have to have 1+1=2 units of energy. Their amplitudes add to 0, so together they have 0 units of energy.
Best Answer
The thickness of the AR coating is chosen such that the reflections from the two interfaces cancel out (at the wavelength for which the AR coating was designed):
See Anti-reflective coating in Wikipedia.
As endolith points out in the comments, to explain how the transmission is enhanced, you have to draw a few more rays in the diagram. Here's another illustration, from the Wikipedia article for Fabry–Pérot interferometer, which shows a few higher-order reflections:
For the anti-reflective coating, you choose the thickness such that R1 and R2 cancel while T1 and T2 constructively interfere. Note that this is dependent on the wavelength, the angle of incidence, and the index of refraction of whatever is being coated. With other thicknesses, you can make a high-reflectivity coating, or a coating of whatever reflectivity you want.