Any power the stator is producing that is not being used (loading up the system will reduce stator voltage) will have to be dumped by R/R and the only way it can dump this is by means of heat.
Spot on
Also as R1ffRAff has said many times any corrosion on R/R terminals will cause higher resistance and turn current flowing through the terminals into heat. This is a vicious circle = the hotter the terminals get the more they corrode - the more they corrode the higher the resistance gets and the hotter they get,
Also spot on
this heat is conducted back into R/R and its components
The increased current draw from the R/R will result in less wasted. Generally; as you've said the more current to waste the hotter the R/R runs meaning a higher current draw will result in a cooler R/R.
The amount of heat "conducted" back through a vehicle wiring system is so miniscule it's not worth mentioning.....
A bulb may be a bad example because they are non-linear and resistance changes as filament temperature changes
Correct, as the temperature increases the resistance increases, just as with wiring, starter motors, coil packs, pumps even resistors
The reason this was used as an example is because in real world scenarios we will rarely see a random resistor drawing that many watts apart from heater blower motors, bulbs are very common for all the obvious reasons
- when power is first applied a filament bulbs draws an 'inrush current' many times greater than its running current, this is because with a cold filament the resistance is much lower than with a hot filament (reason why most bulbs blow when power is first switched on rather than when they are hot).
Yup, it's called peak current draw and then nominal current draw.
In my experience however, (600-1000 miles a week in a lorry) bulbs tend to fail due to harsh bumps....hot filament, quick jolt from a bump and the almost molten tungsten let's go.
The reason why a filament bulb in your house can trip the circuit breaker when it blows is that once the filament breaks it causes an electric arc to form, the resistance of the arc is very low compared to filament and so a lot of current can flow.
AC and DC work differently, with an AC system we deal with higher voltages in order to use solid core wherever possible. Electrons move over the outside of conductors with only a few exceptions and so the surface area is greatly reduced, thus can handle a lower current.
AC has more of a tendancy to arc due to the magnetic field created, I won't go too far into this as it isn't relevant to the OP, the first discharge lamps were DC and VERY unreliable due to the high current required to create the arc, AC ballasts are far more reliable and so the lower current draw is far more forgiving on it's components, especially within the discharge capsule.
If the 55 watt bulb is replace with a standard resistor with fairly flat resistance / temperature characteristics
I presume you mean using a resistor with 5% tolerance.....one with a gold band?
then as the voltage increases the current flowing will increase in direct proportion to applied voltage.
The current will only increase if there is a current draw......I.e. a bulb, motor, heating element ect. Increasing the voltage will only serve to apply more or less voltage to the device in question.
Once you quote a wattage then things change because a 55watt bulb designed to work at 6v will have to have half the resistance to allow double the current to flow than the same bulb designed to work at 12volts.
Spot on
If a connector gets corroded it will cause higher resistance which will try to make current drop, but the problem is that as the voltage across the connector increases this will increase the heating effect on that connector, which with a very small area to dissipate the temperature quickly builds up, and flows back into parts of RR that do not contact heatsink, if added to this the RR is trying to compensate by maintaining a certain voltage at its terminals and voltage drop across external connector means RR will try to compensate by increasing voltage.
You had so much right and changed your mind for the summary 
If a connector gets corroded it will cause a higher resistance.....this will drop the VOLTAGE at the load which will increase the current drawn......please see ohms law for ref.
As the current increases this will increase the temperature of the corroded connector and further increase the resistance.
The amount of heat that can be physically conducted back to the R/R is miniscule. I won't say it can't be conducted back but if I heat just the tip of a length of say 14AWG wire to solder to, or get it cherry red.....20CM away from that point will be ambient or close to. Either way it's negligible. Consider that diodes are generally ok just over 70 deg c, if they weren't then the processor I'm your computer would fail about 10 seconds after switching on.
Having said that, as the old proverb goes. A candle that burns twice as bright lasts only half as long; yes, the hotter something runs, it's working lifespan will be reduced but we're talking about an R/R it's designed to run that hot.
The R/R is doesn't try to compensate, if the voltage drops below 13.5vdc with the lamps on high beam at 5000rpm then it's out of spec; or the charge systems components are unsuitable for the current load required. This is the reason the stator was upgraded on the injection bikes to circa 300w IIRC ...... Its got more electrics to run and therefore more load. The only thing it will do is stop feeding so much current and waste this off as heat.
I lost the quote bubbles somewhere along the way so I've underlined my annotations, perhaps if I ask very nicely....a mod might clean it up for me
SV650rules I hope this helps you as well as the OP or anyone in the future reading this thread.