[Platinum]

 ;)I am purposing on an oxidative cleaving of an aliphatic carbon-carbon double bond. According to all sources this far, only the oxidation of a carbon-hydrogen without cleavage (a carbon at the end of a C-C chain) to an aldehyde, two electrons are withdrawn - this is correct?

Now, if the chain is supposed to be broken at a double bond, where a cleavage occurs and oxygen replaces that C-C double bond, does it takes four electrons to perform this? I am thinking like oxidation agent pulls two electrons to remove the double bond, and pull two electrons more from the carbon which oxygen binds to. Is this correct?

An instance is below:

[Platinum]

Many scientific reports and works do the oxidative cleavage, but they never really explain how many electrons it takes. One source mentioned that they used 1:2 ratio, where 2 was the molar ratio of H2O2. Because 1 mole H2O2 is withdrawing 2 electrons, I did interpret this as when they are using 2 moles H2O2 for every mole of substrate, it's about 4 electrons totally required in the process.

[Babcock_Hall]

Why don't you assign oxidation numbers to the carbon atoms undergoing reaction and see how they change?  This is a bookkeeping device, but IMO it is a useful one.  The oxidation number of a carbon atom = (number of bonds to O, N, Br, Cl, etc.) - (number of bonds to hydrogen).  If a carbon atom changes its oxidation number from +1 to +2, for example, it has formally lost one electron.