If you think about it, all you do is add a carbon, so what you need is a nucleophile that can attack at the carbonyl carbon that looks like an $\text{R}$ group.

A straightforward solution is $\text{R}-\text{MgBr}$, an alkyl Grignard reagent that acts like an alkyl anionic nucleophile.

where $R{}_1$ and ${\text{R}}_2$ merely differentiate between unique $\text{R}$ groups. The only problem is that the tetrahedral intermediate, after being protonated, no longer possesses a carbonyl group. So you have to oxidize it to turn it back into one.

Overall:

1. Attack the carbonyl with an alkyl Grignard reagent to tack on the alkyl group.
2. The acid protonates the alkoxide to finish the first step, and the water that remains deactivates the alkyl Grignard reagent back into an alkane.
3. PCC (pyridinium chlorochromate) oxidizes the hydroxyl group into a carbonyl group via a $\beta$-elimination of the explicit proton.

PCC looks like this:

What's special about it is that it oxidizes alcohols one step forward, i.e. primary alcohol to aldehyde, or secondary alcohol to ketone.

One way you could do it is using trifluoroperoxyacetic acid (a Baeyer-Villiger oxidation), and then using diisobutylaluminium hydride (DIBAH) in Toluene in the presence of a dry ice bath.  take acetone and convert it to acetaldehyde:

Neither mechanism is particularly well-known, but in general, in step 1, an oxygen is inserted in between a carbonyl and an adjacent carbon, and then for step 2, the $-\text{O}-\text{R}$ group is reduced to $\text{H}$.