I am going to regret this......
OK, first of all, I am in awe of your fabricating skills and your patience.
You have created a truly fantastic setup.
However, you are way out in the bushes when it comes to the electronics.
I am an electronics engineer, with 30 some odd years of working on stuff like this.
I say this, not to brag, but assure you I know what I am talking about.
I am trying to not come off heavy handed, ask around, I get along with everybody. But you have several major flaws in your electricals.
First, your power blocks. Are those steel? They looked like it when you were building them.
Really really should have been made out of copper.
Silver would have been better, but would be totally overkill.
They can be saved though. Just make a thick copper strip to go along the top of the steel part.
Next, basic law of series DC circuits:
Current level is the same at all parts of the circuit.
ANYWHERE you break that circuit you are switching the exact same number of amps.
So ground switching is useless.
Next, any brush-type dc motor is also a dc generator. And vice-versa. I have seen old type generators squeal the belt if the points in the voltage regulator stuck when the motor was switched off.
I do not know of the fan motors in question are brush-type DC, or AC.
Given that they seem to be generating power when spun, I'm guessing DC.
And, before anyone jumps me, yes, they are going to AC motors in cars to a small extant. Not so much with the big motors like these.
Brush-type DC motors wear out and make electrical noise.
Plus they are inefficient.
With cheap solid-state equipment now, its easy to use a brushless AC motor on DC.
You just build an oscillator that converts the DC to AC.
All the 12 volt fans in modern computers are brushless AC.
As for the protection diode you first had in the circuit, all it would do would be to drop .7 volts going to the fan.
You need a protection diode across the coil reverse wired to stop what is know as back emf.
That occurs when the armature of the relay moves in the collapsing magnetic field of the coil.
Probably not needed in this case, and may be built in to the relay anyway depending on which model you have.
Now, the circuit problems you are having.......Dear God, whats with those switches??
You are showing a spdt switch.
It looks like you have the anode of the indicator diode on the common, along with terminal 85.
I assume that the switch is spdt, not spdt-center off.
When the switch is in one position, the anode and terminal 85 are hooked to ignition power.
But, when you switch it the other way, you ground those two terminals.....Why?
OK, just looked at that switch again, and I realized it must be a lighted switch. The way you show it is OK, but just a little misleading.
I am guessing now, its a spst. Not spdt. Makes a little more sense.
Remember, you do not know what is in those LEDs.
Since they run on direct 12 volt, and will compensate for voltage fluctuation, they do not have simple resistors inline.
As you discovered, that will not work.
When I build a 12 LED setup for someone, depending on who I am building it for, I may get creative.
They make LED driver chips that will take a wide variety of voltages in and output the correct voltage and currant for the LED you are using.
I often use one of those with a full wave diode bridge.
By using one of those, it does not matter which way they hook it up, the polarity will always be correct.
When you turn the switch off, of course, you are going to reverse bias the LED and blow it if its not protected.
The LED has to see a voltage DIFFERENCE across it in the correct direction to light up.
When the switch is on, but the coolant switch is off, the LED is seeing 12 volts on both sides of it.
Ignition switched 12v on one side, and 12 volt through the fan from the terminal block on the other.
What you have created here is the equivalent of a ground loop.
Remember I mentioned those terminal strips needed to be copper?
What is happening here is that something on the positive strip is drawing current, and there is enough resistance either in the block, or somewhere between it and the battery to cause a voltage drop.
So that the ignition 12 source is a volt or two higher than the terminal strip.
Which lights up the diode.
Or the fan rotating is generating voltage and back feeding the strip.
There are so many possibilities here, I'm not sure I can isolate the correct one without being there.
If you have a multimeter, and want to get to the bottom of this, let me know and I can probably talk you through some measurements that will solve it.
Its been a really long day, and I fell asleep twice typing this, so if it isn't clear, let me know and I will try to fix it tomorrow.
J.