Here is a relevant extract from a PDF (Dr. P.G. Hultin, 28 July 2016) depicting oxidative cleavage. For the compound in hand (your question) the process will be done twice. $\ce{H2O2}$ will serve as the source of $\ce{H3O+}$.
When alkenes are treated with $\ce{KMnO4}$ in acidic solutions, the diol is not formed. Instead, the alkene is cleaved. The reaction proceeds by the same mechanism at the start, forming a cyclic manganate ester (although since the reaction is under acidic conditions the structure is protonated). In the schemes below, the alkene carbons are highlighted throughout, so you can see where they end up in the product.
Now, under these conditions the manganate ester is not very stable and it undergoes a cyclic fragmentation process, which results in breaking the C-C bond between the two oxygens. Notice that in this case, since there was a hydrogen atom attached to each of the alkene carbons in the starting material, there is a hydrogen attached to the carbonyl carbon in the product and therefore the product that is initially formed is an aldehyde
Aldehydes are very easily oxidized to carboxylic acids, and thus the aldehydes formed in the cleavage reaction do not survive. They are rapidly transformed into carboxylic acid groups, by a complex reaction whose mechanism you need not worry about.
Now, if the alkene had not had any hydrogens attached, the product in that case would have been a ketone
rather than an aldehyde. Ketones are not easily oxidized further, and the reaction would have stopped at that
stage.
If one of the alkene carbons had a hydrogen substituent, while the other did not, then we would get both acid
and ketone groups in our product, as shown below.
