Electronic – Why voltage decrease in battery when more current is drawn from it

According to the BQ2002N datasheet the battery connection schematic is:

The datasheet says that RBAT Battery input impedance is 50M Ohm so I don't see why you selected such low resistance values (47R + 10R) that actually drain your battery with 120mA consumption.

In addition according to the datasheet:

Note: This resistor-divider network input impedance to end-to-end should be at least 200kΩ and less than 1MΩ.

You should use much higher values like 499k and 100k (1% series).

You have connected the CC output of the IC to the battery but that can't work because that pin is An open-drain output used to control the charging current to the battery. CC switching to high impedance(Z) enables charging current to flow, and low to inhibit charging current.
So basically the pin can only be pulled down or float, it can't provide power to the positive pole of the battery.

I would suggest you follow the schematic shown in DV2002L2/TL2 Fast Charge Development System.

12V,1Ah means you've got 12Wh. 14V,1Ah will get you 14Wh.

So yes - you've got more "power" (and you did increase the capacity of the batteries by that...)

Now it depends on how your circuit will use that energy. When it needs less than 0.1Ah @ 14V, it will work longer. If it "burns" the excess voltage, and still need 0.1Ah, regardless of whether it is running at 8,12 or 14V, it will run for the same duration