No clarification, no answers from anyone else who is helping. OK, so this is my GUESS.
Lipo cell ->charge controller-> boost converter->12v load and I'd like some help.
Since other smart people are answering this like nothing's wrong, I feel like I am wrong.
But I would expect this to work a hell of a lot better:
Lipo cell > X > IN boost converter OUT > 12.8V output
External power supply > IN charging circuit OUT > X
The LiPo cell, charging circuit output, and the boost converter input are all connected together. (This is why there are two pairs of solder pads in the charge circuit board on the output side, marked as the Battery +- and the Output+-). The voltage you measure on any one of these 3 things will be the same. And it should be something like 3.5 to 4.2V, depending if the radio is on/off and how charged up the lipo is. You cannot measure 4V on the lipo and 1.1V on the charge circuit output unless you have it hooked up in some way I am curious to understand the purpose of. If you cannot draw up a schematic, maybe you can at least tell me what is the charge circuit? Why did you buy it, and what do you expect it to do? It's definitely not going to charge your lipo if you connect the battery to the INput and have no external power source in sight.
[edited for errors]
*Ok, just looked at your link to peek at this board. kjr18 suggests this board might be used for output protection. And looking at the board, he is correct. (But this doesn't change the way you connect it. Still gotta do it my way).
The big chip is the charge IC. The tiny SOT is a voltage detector. The medium chip is the output FET/switch.
The battery goes on the B+- pads. It looks like the B+ and Out+ are in continuity. The FET looky thing is next to the ground pads, and it would be an NFET, drain on the OUT- and source on the B-. When the voltage goes under the specified level, the detector switches off the NFET and saves the day.
But you have soldered your battery to the Input pads. The input pads will likely be directly connected to the USB connector power and ground. That would be so you could hardwire a 5V power supply in lieu of using the USB connector. So if you plug it into a USB port in this state, and if the power source doesn't cut out immediately, you will damage your cell.
This explains your readings. You are powering the battery charger input with the battery and using the charger output to feed the boost converter. The IC detects no voltage on the battery and initiates trickle charge. This kicks the boost circuit on and gives the radio just enough juice to malfunction. And the resulting voltage at the output of the charger/boost input is 1.1V. Which is not enough to power the voltage detector, either.