However, I am really not that good at EE, so I am clueless as to what the unlabeled resister values would have to be in order for this circuit to function properly. I realize that there are protection circuit ICs, but I want to build the actual circuit myself. Does anyone have any insight as to how this circuit works and what the resistor values would need to be?
I would suggest using TP4056 IC for proper 3.7v battery management system
Here is an out-of-head blockish diagram of the minimum that is required to do what you ask for. This is not a complete charger. It would prevent battery charging above some Vlimit and prevent discharging by load below some limit.
It applies Vcharge whenever the battery is below some chosen voltage.
It supplied LOAD from LiPo as long as Vbattery is above some chosen voltage.
It does NOT ipmplement CC/CV charging.
It does not implement low low battery lockout or low battery trickle up.
R3 R4 set high voltage battery cutoff.
Vstop_charging = Vref_D1 x (R3+R4)/R3.
R5 R6 set low voltage output cutoff voltage.
Vout_stop_working = Vref_D2 x (R5+R6)/R6.
D1 D2 = TLV431 . Vref = 1.25V.
R1 is a pull-up/off for Q1.
R8 limits bse current for Q1.
R2 turns on M1 when Q1 is not turning it off.
R7 turns of M3 when D2 is not turning it on.
R3 R4 R5 R6 discharge the battery at all times and need to be set as high as possible. See datasheet.
The diodes inc TLV431 is probably superior to most.
The shunt regulators operate like overvoltage switches here, if the voltage at the reference pin is higher than their internal reference then they conduct.
So the first one is the overvoltage detector and so the voltage divider on it should exceed the internal reference at the upper voltage limit. When conducting it pulls down the input to the second regulator which prevents it from conducting.
The second regulator conducts when the voltage is above the low limit and turns on the NPN on the output.
The diode and components behind it seem to be for turning the NPN on for a short time while that small capacitor charges, I am not sure why this is done but may be to remove a small overcharge from the battery, you can probably take it out but I would leave it in since it is probably there for a reason.
I am guessing you can calculate the labelled resistors but as for the other ones the base pull-down can be any large resistance, try 10k. The resistors on the capacitor will change the initial turn-on current and how long it stays on for, not really sure about this one but 1k for the top resistor and 100k for the bottom one seems reasonable. Start with a capacitor of 10uF but if you have problems with it locking out on a slightly overcharged battery then you may need to increase it.
Also if you are drawing more than an amp or so you will want to replace the NPN with an NMOS transistor since they have much lower resistance for a small gate signal, just make sure the turn-on voltage is above the low voltage limit.