How does LED Flashes One by One in my Project Some Queries?
The overwhelming problems here are that the circuit diagram is wrong and cannot work
 AND the circuit he has built DOES actually match the diagram!

Nowhere in the video (I skimmed it) does it show the flasher flashing - which would seem somewhat strange.   

While this claim seems "rather hard to believe" [tm] a look at the many many many 2-transistor astable multivibrator circuits from the link below will show that they all work the same way AND that none match his circuit. 

The problem with his circuit is that R3 &amp; R4 should go from the base of each transistor to B+ (battery+) and NOT from base to ground.

Here are many many many circuits that agree with each other and which his fails to match - 



Here is the rear view of his DED Flasher which shows that it is indeed built as he shows in the circuit diagram:

This circuit diagram from here



Or this one from here



Or all the following and more from here (as above) show he is wrong.

Many of those pages will describe correct operation BUT before you read them, with the new circuit, see if you can see how it works, then look it up. The correct answer will almost certainly surprise you. 





OPERATION:

Assume no cap leakage.

All off.
 C1! discharged,. C2 discharged.
 Power on.
 C1+, C2+ pulled high via LEDs by battery.
 As C1 and C2 have no V across them the -ve sides will also be pulled high.
 Bothe bases pulled high - boty transistors try to turn on.
 One will win.
 Say Q1 turning on faster:
 C2 will have some charge as -ve side discharged via base (or left side chgd via R1 D1).
 QA Q1-c is driven to ground, C2- will be driven to BELOW ground.
 Q2 driven hard off
 If C2- driven more than about 0.6V below ground D4 conducts to clamp -ve excursion.  Also provides a discharge path via Q1 - see below.
. 
 Now Q1 on Q2 off.
 C1+ will charge via R2D2 (not a robot).
 Usually R4 is to B+ and it charges C2 so Q2 turns on. 
 Q1 is on as above as C1 high but C1 is charging via R2 D2 C1 R3 so -ve end falls until Q1 is not held on as Vb too low.
 But now C2 is discharged due to path Ground-D4-C2-Q1. So when Q1 turns off Collector of C1 rises and lifts C2+ and so C2- follows so Q2 turns on and identical second half cycle happens 

This is a much damaged version of the original as usually bases are driven negative and do not conduct during timing part of cycle and resistor charges capacitor via a normal time constant. Here the caps forward bias the bases when the opposite collector rises and you get massive base current pulse via eg R1 D1 C2 Q2_base. So time constant of RC is much shortened so oscillator will run rather faster than usually. T = RC of bases is notionally 10k x 100 uf = 1 second but because capacitors are clamped and discharged directly into bases this will run much faster. 

So it works after a fashion but is crippled.