Let me familiarize you with ohm's law.
R=E/I, or I=E/R, or E=I*R. These are all the same formula, simply rearranged. Where E= Voltage, I= Current, and R= Resistance.
In this case, you are searching for voltage loss at some sort of junction, be it switch, connector, or fuse clip.
We are searching for voltage loss between two points, or E, so use the formula E=I*R. Notice if I = 0, any value of R has absolutely no effect on E. What this means is that to measure resistance in an active circuit, you must have both current and voltage applied.
Effective test methods:
The voltmeter can directly measure the voltage potential difference between its probe tips. So, if you place one probe on the battery POS terminal and any point such as at the bulb's power in terminals, it tells you directly how much voltage was lost in the routed pathway between those two points. This test method works for all individual connection points and switches which has current flowing between them. However, the voltage reading is directly proportional to the current that is flowing through the switch or connection at the time taking the voltage reading. Zero current flow will always show a zero voltage loss along a pathway that is only distributing voltage, regardless of actual resistance (short of infinity).
Say you wanted to test a switch for contact resistance. A DMM's ohmmeter function induces such a tiny current into the circuit being tested, that low resistances are difficult to register accurately.
Back to the equations: If I =1 and R = 0.5, the voltage read across the resistance is .5V. However, if I =10 and R is still 0.5, the voltage dropped across the resistance is 5V!
You state :
As I mentioned, earlier in this restoration I fully disassembled the left and right switches and renewed them properly -- when I was done, there was nearly no resistance on the paths through the switches themselves. They flowed current in a truly excellent fashion.
While this is a lovely statement, it has no solid data numbers with which to verify the supposition, or any test conditions, parameters to frame the test report within.
If we work with such fuzzy data as fact, we will be as baffled as you. Part of being a technician is the ability to set up meaningful test conditions.
Frame ground is NOT the primary ground path normally used on the bike, apart from the starter. ALL other devices use the harness green wires to reference Battery NEG terminal. You CAN cobble something to work without using the green wire distribution. But, someone with electrical knowledge will consider it a hack.
Every single headlight has a wattage rating. I don't understand why you think you can ignore this important parameter, as it directly impacts both voltage and current use on the bike.
The same headlight without the relays was obviously dimmer (aimed at my garage wally) than that headlight with the relays.
As I explained earlier, this is because you failed to find out exactly where resistance has been inserted between headlight and battery. Your relays have simply bypassed the the problem you are ignoring. I can assure you that Honda's headlight was not dim when new, with or without higher wattage headlights. There is no designed in reason why your circuit modifications are necessary, unless you have replaced the connectors and switches with inferior devices.
Beyond this, I guess I really can't help anymore with your project, as I now have absolutely no idea how your bike's wire circuits have been modified away from Honda's book representation, working or not.
Hopefully someone else can guide you to a satisfactory conclusion.
Good luck with your experiments!