Well, you can make a good guess using oxidation numbers. If you find a species with a very positive oxidation number, it generally is what is being reduced.
If you work out the oxidation number of Mn in MnO4^-, it should be +7, meaning it’s already pretty oxidised and ought to get reduced.
As MnO4^- has all these oxygens bound to it, which it would have to get rid of to be reduced, the oxygens must be removed in the form of water molecules (the acidification of the solution provides the H^+ ions required for this). You have to memorise the fact that the manganese is reduced to Mn^2+ at A level.
C2O4^2- is a little more tricky. Because you’ve worked out that the MnO4^- is getting reduced, then C2O4^2- must therefore be oxidised. That means it must form electrons as one of the products. Although C2O4^2- has several oxygens on it, it doesn’t need to remove them as water molecules (because this isn’t reduction). It can instead break down to form CO2.
Can you now construct some balanced ionic equations for each process?