LTC4353 secondary input voltage drop
I think the problem is with your logic. If you go to the datasheet, on page 5, this is what it says:



With the voltage divider you currently have, it's not possible to enable the MOSFET which switches on +BATT (Q1 on your schematic).

You have a divider ratio of:

$$\dfrac{165k\Omega}{165k\Omega+10k\Omega} \approx 0.94 $$

(Are the resistors swapped? This could fix the issue)

So whatever your input voltage is (Vin) gets scaled by that and depending on whether that value is greater than 0.6V, you can enable or not the +BATT MOSFET. So, even if Vin drops to 3.3V, you still can't turn that on (would have about \$3.3V(0.94)=3.1V\$ at the enable pin—far away from turning on the gate since it needs to be \$\leq 0.6V\$).

The reason why you see a voltage drop of about 0.6V is because of the body diode. Since +BATT in your scenario is greater than Vin, +BATT still drives the output but instead of having a path through RDSon of the corresponding transistor, it does it through the body diode of your MOSFET.

I'll show you what is happening:







simulate this circuit &ndash; Schematic created using CircuitLab

Hopefully this helps.