Learn me turn signals?

A friend of mine brought his new-to-him 1997 Ford F250HD by the shop last night.  During his Sheriff's Inspection, the turn signals and the hazard lights just stopped working and he couldn't figure out why.  He had replaced the round flasher unit in the fuse box and checked the appropriate fuses, yet he still had nothing.  I back-probed the wires that were supposed to carry the +12V for the turn signals and the hazard lights and got power on the hazards but not on the turns. Still no blinking lights of any kind. 

Thinking that maybe there was something loose in the harness, I reached up behind the dash and started wiggling things around.  Doing that, I discovered a second flasher unit on the back side of the fuse box and it felt a bit warm.  We put a new flasher in the back and now everything works as it should.

The owners' manual and the fuse box cover diagram both show only one flasher.  I get how the flasher works and that the power comes from the PDC under the hood into the fuses for the turns and the hazard, and then to the flasher, then up to the multi-function switch, and then to the lights.  But why two flashers?  And if there's one for the turns and one for the hazards, why did neither the turns nor the hazards work when only one of them was failed?

It's all fixed now, but I don't understand how or why....  Clue me in?

Turn signals, IIRC, go through the hazard, so if the hazard side dies, the turns do too.

SOP on pre-BCM GMs with inop turn signals is to manipulate the hazard switch to see if that fixes it.  Have done many hazard switches.

"Learn me turn signals"

My first thought: "Do you drive a BMW?"

Yeah, I was thinking the same thing when I posted!

I can sort of see why they would have separate flasher units for hazards and turns but if one is dependent on the other, that seems like it defeats the purpose.  If the truck was still in front of me, I might play with some continuity checks to see if I could figure out how it was wired.  Does the output side of one flasher feed the input of the other?  It would almost have to, wouldn't it?  And then how does that work?  You'd have the resistor on the first heating up and breaking the circuit first and the second would never get there?  Except when you're powering four lights instead of two the current draw would be higher.... But then you'd still expect that only one of them would break, right?

I'm still lost as to why it's set up that way and how that would work better or be cheaper than a single flashing unit.