Determine oxidation in organic chemistry through oxidation numbers?
Yes, determining oxidation numbers and thereby finding out what is oxidised and what is reduced is a valid method, especially for reactions that include oxidations or reductions.

Your ‘problem’ with an aldehyde and a ketone is actually not a problem at all. Ketones have an oxidation state of $\mathrm{+II}$, which means that the only thing they could be oxidised to in a two-electron (i.e. non-radical) process would be $\ce{CO2}$ — hence why they are stable towards oxidation. Aldehydes, on the other hand, have an oxidation state of $\mathrm{+I}$, which means that they can easily be oxidised (to carboxylic acids, oxidation state $\mathrm{+III}$) by simplet two-electron processes.

Likewise, the different oxidation states of primary ($\mathrm{-I}$), secondary ($\mathrm{\pm 0}$) and tertiary ($\mathrm{+I}$) alcohols can be used to help understand why it makes sense to distinguish.

However, do note that many organic reactions — while definitely involving a change of oxidation states — do not fit into the classical redox paradigm well. Throughout the mechanism of a Horner-Wadsworth-Emmons reaction, the (tertiary) phosphonate carbon changes its oxidation state from $\mathrm{-II}$ to $\pm 0$ in two distinct, one-electron steps, while the aldehyde carbon goes from $\mathrm{+I}$ to $\mathrm{-I}$ also in two distinct steps — but nobody would go as far as calling it a redox reaction.