The coating has nothing to do with it - it's the metal of the pan that matters. As was mentioned in the comments, many Teflon coated pans are made of aluminum (or aluminium, depending on where you live...). The issue is skin depth: for a non-ferromagnetic material, at the frequencies of the induction heater a large fraction of the volume of the pan becomes a (good) conductor of the electrical current: and because the size of the (eddy) current is fixed by the geometry of the cooker, if you present it with a low resistance, the amount of power generated will be low, since $P=I^2 R$.
The following table of skin depths (at 24 kHz) shows you what is going on (adapted from http://en.wikipedia.org/wiki/Induction_cooking):
rho mu skin R effect
Copper 0.68 1 0.017 0.04 1.00
Aluminum 1.12 1 0.022 0.051 1.28
Stainless steel 304 29 1 0.112 0.26 6.5
Carbon steel 1010 9 200 0.004 2.25 56.25
Stainless steel 432 24.5 200 0.007 3.5 87.5
in this table
rho = resistivity (in 10^-6 ohm-inches)
mu = relative permeability
skin = skin depth in inches
R = surface resistivity (in 10-3 ohm/square)
effect = relative heating effectiveness compared to copper
As you can see, copper and aluminum make terrible materials for induction cooking pans: you need something with high electrical and high magnetic permeability to get the best heat transfer. 432 Stainless (which has high resistivity AND is ferromagnetic) is a lot better. Some copper pans adapted for induction heating have a coating of another material specifically for this reason.
A better term than "coagulation" is "denaturation". To denature a protein is to unfold it, and this can be done by thermal energy. Initially the egg white proteins are each folded up into little ball-like bundles which limits their interactions, but when they denature the become strands. The charges on these strands then attract to each other and form a tangled mess. Of course, this changes the entropy, but not as much as for simple molecules. That is, we need to look up the "heat of denaturization" for albumin, which is 12.2 J/g (compared to water melting, 334 J/g, so it's fairly small).
That is, denaturing a 60g all albumin egg, should take 60g * 12.2 J/g or 732 J. Given the heat capacity of water as 4200 J/kg K, or 168 J/K for 40 mL of water, this should cause a temperate change of about 4.3C in 40 mL.
So the specific answers are:
The heat of denaturation is small but not completely negligible. Since you start with 100C water, and want to end around 65C, for the small amount of water you mention (40 mL), it's around a 10% error.
Yes, if you're trying to hit it right at 63C, you'll need to account for the heat going into denaturation.
For a temperature controlled system like a sous vide cooker, the amount of energy going into denaturation is negligible compared to energy lost to steam from the bath, etc. (Overall though, this question doesn't quite make sense to me and I wonder whether you're confusing "temperature" with "heat".)
Also, keep in mind that denaturing the albumin does not necessarily mean the egg is cooked sufficiently to be safe to eat.
Finally, I wouldn't trust the 63C number, especially since you're taking the low end of a given range. If you really want to be exact, you'll probably need to go to the source of those numbers. (And don't forget the heat capacity of your container, and also I suspect heat loss will, in fact, be significant here.)
Best Answer
Because the electric resistance of a human body is by orders of magnitude higher than the resistance of the steel pot.
According to Maxwell–Faraday equation, changing magnetic field creates the electric field, i.e. the difference of the electric potential. Then, the difference of the electric potential creates electric current. Electric power = voltage * current. The voltage (the difference of the electric potential) only depends on the properties of the changing magnetic field. The current is defined by the Ohm's law. Lower the resistance, more current, more power.
For example, if you're trying to heat a glass pot, the changing magnetic field will still create the same electric field it'd do in a steel pot. But since glass is insulator, no electric current will flow, and (almost: dielectric heating is negligible @ the operating frequencies of induction cookers) no heating will occur.
P.S. Most funny thing will happen if you'll try to heat a super-conductive pot. To be honest, I'm not exactly sure what will happen.