MC34064 Step-Up Regulator Design
I will not describe how a boost converter works, because it's not too difficult to reach to the required information.

But I have to say that the switch (inside the IC) will never turn on thus the device will never operate when the input voltage is higher than the desired output voltage (7.4V input, 7V desired output). Thats why you see 6.8V output (7.4V input minus 0.6V diode drop equals to 6.8V).

Besides, 100uF timing capacitor will never let the IC to operate. You should decrease it to something like a few nano- or even pico-Farads.