Your first circuit is a fine example of how to create a bad and potentially explosive circuit. If the battery is low it gets hit with 3.3V and as much current as the LDO can supply rather than being carefully charged.
When the battery isn't low the charger is powering the circuit rather than the voltage regulator. The charger has no idea how much power is actually going into the battery and the circuit is being powered by up to 4.2V rather than 3.3V.
In the second circuit when VIN is high Q1 is off effectively isolating the battery from the MCU pin allowing the battery charger to do it's job.
Power flows to the MCU pin via a diode, assuming 5V input with an LDO down to 3.3V the 0.2-0.3V drop in the diode shouldn't be an issue.
When VIN is off R2 causes Q1 to switch on and allows battery power to flow from the battery to the MCU pin.
Q1 needs to be sufficiently large to handle the current with appropriate voltage thresholds so that it is fully on and off under the normal operating conditions.
So yes, connect the MCU pin to the input of your regulator and you should get a clean regulated supply until your battery discharges to about 3.5V (assuming a 0.2V drop out LDO). Once your LDO drops out of regulation you should shut the system off since it's no longer reliable. You should definitely shut off before the battery gets below about 3.1V or you risk damage to the battery pack.
3.5V will normally be about 30% charge remaining but it is very temperature dependent. You could hit that voltage at 90% charged in low temperatures.