Your circuit is apparently working fine, you just don't realize it.
With the drain left floating, the MOSFET cannot pull any current, and therefor the integrator keeps providing a lower and lower gate drive, trying to draw current.
With the drain grounded and the reference at 0 volts, the integrator goes just low enough to start pulling a tiny amount of current and satisfies the loop.
What you now need to do is get a set of power resistors, such as 10 ohm / 25W, 20 ohm / 25 watt, etc. and use them as loads. From the voltage across your load resistors you can determine the current through them.
However, before you do that, you need desperately to provide a heatsink for your MOSFET. Its' peak power dissipation at a current of 2A will be nearly 30 watts, worst case (for very low load resistances).
You also need to make sure that your shunt resistor is of high enough power. If you're following the app note religiously and using a .2 ohm resistor, it needs to be a 1-watt or better unit, since at 2 amps it will dissipate (2 x 2 x .2) = .8 watts.
You would also do well to reconsider your power wiring. The contacts on a solderless breadboard may not handle 2 amps well.
Finally, you need to make damn sure you can't provide a reference voltage greater than 2 volts.
And finally finally, I suggest you do a good deal of testing with resistors, in order to convince yourself that the circuit really is working. I suspect that you may find it doesn't work nearly as well when you give it a highly inductive load like a coil. It may work, and it may not, since the effects of the inductance will tend to counteract the effects of the loop integrating capacitor, and the result may be instability.
Otherwise, congratulations.
I will answer your questions in the order that you asked them.
1) First you need to make sure you have a 470uF to 1,000uF capacitor from the power pin to ground pin. That will get rid of the 'self-oscillation' effect.
2) The .05uF and 10 ohm resistor on the output act as a phase-shifter, to cancel out the internal phase shifting caused by the LM386's own design. It is common to see this on many IC amplifiers.
3) The capacitor in series with the output not only blocks any DC voltage from going through the speaker, it effectively creates a new 'AC' based signal, with no DC component.
4) Because there is a DC voltage at the output equal to 1/2 your power supply, there has to be a DC "blocking' capacitor with a high enough value to allow some bass to come through.
5) You can let the bypass pin float, but a 10uF capacitor to ground (the pin gets the + lead) may help with noise. Use it only if needed.
Best Answer
Lets see what we're trying to implement here:
We get the desired results by the following means :
From the first glance it looks like \$Z_{in}=2M\$. Well, you could achieve the same without bootstrapping, so why bother? It turns out that the input impedance of this follower is higher than 2M. In fact, the input impedance in this configuration will (usually) be determined by amplifier's input impedance, and not by the values of voltage divider resistors.
Lets see how it works:
DC is blocked, therefore all the signals in this explanation are AC at the frequencies of interest.
Step-by-step propagation of signals:
The only possible steady state in this condition is: \$v_+ = v_- = v_{out}\$.
Up until now I described the basic follower. However, in this configuration there is one addition - \$v_-\$ is also fed back to the upper resistor's second terminal (lets call this resistor \$R_1\$). So, what the voltage at the resistor will be then? In terms of amplifier's inputs it will be: $$v_{R_1} = v_+ - v_-$$ From the above discussion you may think that this voltage will be zero, but let's not forget that the above discussion is the idealized case. In fact, the gain of the follower won't be exactly 1 - it will be very close to 1, but not 1. What are the implications of this? Well, the voltage on the resistor will be: $$V_{R_1}=v_+ - v_-=v_{in}-Av_{in}=(1-A)v_{in}$$ Since A is very close to unity, the voltage on the resistor will be almost zero, but not zero. This voltage will give rise to current of magnitude: $$I_{R_1}=\frac{(1-A)v_{in}}{R_1}$$ What is this current? It is a current drawn from the source by the resistive divider! If we try to express the "effective resistance" of the divider by comparing the above equation to Ohm's law, we will get: $$R_{eff_{divider}} = \frac{R_1}{(1-A)} >> R_1$$ Given that this huge effective resistance of the divider is in parallel to the input impedance of the amplifier (which is big, but not that big), you can understand why the input resistance of the whole circuit will be determined by the latter.
An answer: You may notice that I did not answer your question yet. Well, the answer to your question is that a tiny current source is just a way of modeling the tiny current which will flow trough \$R_1\$. It is not really zero, but very close to it. Don't take this too seriously - if you understood the whole explanation I wrote above, don't care about not understanding some alternative (and useless in my opinion) way of describing this circuit.