Counting electrons in Ozonolysis
If you want to demonstrate that the addition of ozone to an alkene is a 6-electron oxidation of the carbon framework, here are two methods. To assign a charge to the carbon atoms of a neutral compound (sum of all charges equals zero) treat all hydrogens as +1 and all oxygens as -2, then do the math. Each carbon in ethylene (1) has a calculated (in red) value of -2. Molozonide 2 has carbons with a charge of +1 but, for convenience, they are written as zero and +2. Already one can see that 1 $\rightarrow$ 2 is a 6-electron oxidation [-4e $\rightarrow$ +2e]. Collapse of the molozonide gives formaldehyde (3) of carbon charge zero and zwitterion 4 with carbon charge +2. These species recombine to form ozonide 7 that has the same carbon charge as the molozonide 2. That the zwitterion 4 is the equivalent of formic acid 6, follow the steps 4 $\rightarrow$ 5 $\rightarrow$ 6. Thus, both the molozonide 2 and ozonide 7 are equivalent to one formaldehyde and one formic acid.
Alternatively, oxidation of ethylene (1) to ethylene glycol (8) [-4e $\rightarrow$ -2e] is a 2-electron oxidation and a second 2-electron  oxidation [-2e $\rightarrow$ 0e] affords two moles of formaldehyde. Oxidation of one of the formaldehydes to formic acid [0e $\rightarrow$ +2e) is the third 2-electron oxidation. Both pathways demonstrate that the oxidation of ethylene to formaldehyde and formic acid are 6-electron oxidations.