Need help in understanding the working of battery level indicator
Imagine that all the NPN transistors behave as a controlled switch. The switch contacts are between collector and emitter. When the switch closes, the LED turns on.

That switch is controlled by the Base-Emitter voltage \$V_{BE}\$. \$V_{BE}\$ needs to be more than 0.7 V for the switch to close and the LED to turn on. That \$V_{BE}\$ input of the transistor behaves very similar to a diode (it is a diode!). This means that when \$V_{BE}\$ &lt; 0.7 V there is (almost) no current flowing between Base and Emitter. When more than 0.7 V is applied the Base-Emitter operates in forward mode and it just "eats up" all current that is fed to it. This means the voltage will stay close to 0.7 V.

Note how all the Base-Emitter connections are all in series. If nothing else was connected then to make all 5 \$V_{BE}\$ = 0.7 we would need to apply 5 x 0.7 V = 3.5 V.

But there are some 10 kohm resistors as well. I will call the transistors Q1 to Q5 from left to right. The one on the right will pull up the voltage on the Base of Q5 (on the far right). This means that Q5 will be the first to see 0.7 V across its Base-Emitter when the battery voltage increases. 

As the battery voltage further increases the \$V_{BE}\$ of Q5 will stay 0.7 V. In a similar fashion the other 10 kohm resistors will apply voltage to the other Base-Emitters. As the Base of Q5 is 0.7 V, there will need to be 1.4 V at the base of Q4 before it will start to conduct. In the same way Q3 will need 2.1 V at its Base. So from right to left each transistor will need a higher battery voltage to start conducting.