A) How does the internal field result in the progressive voltage drop (relative to the down-stream terminal) along the wire - what is the mechanism? Is it to do with a changing concentration of electrons?
Imagine a rock rolling down a hill. The rock is moving from high potential (top of the hill) to low potential (bottom of the hill). We say the interaction between the rock and the gravitational field does this. The field does work on the rock as it falls.
The same thing happens in the resistor. The only difference is that instead of the gravitational field being mostly static with near constant strength in the vicinity of the earth's surface, the field in the resistor is very dynamic. The progressive voltage drop only happens once a field is established within.
The field is a result of the arrangements of charges. As the charges are mobile in a conductor (such as the wires and the resistor), these fields are dynamic and change as conditions (such as applied voltages) change.
B) why isn't the internal field zero, as in a closed metal container?
Because there is a power source providing charge and maintaining the field. A complex set of interconnected water pipes will tend to have the same level of water at all places if the water is static. But if you start pouring in water on one side, it may pile up there because the water in other locations takes time to move out of the way.
Without power, the charges would migrate away from the interior of the conductors, leaving a zero field inside.
But the voltage source is able to provide charges onto the wire. The charges migrate to resistor. At first, there is no field inside the resistor, so the charges inside are initially at rest. But as excess charge arrives at the resistor, it begins to build up. This quantity of excess charge induces a field inside the resistor. This field moves charges within. This process continues until the quantity arriving at the resistor exactly equals the quantity departing the resistor.
