Number of moles of MnO₄⁻ required to oxidize one mole of ferrous oxalate completely in acidic medium will be :

Reduction half reaction in acidic medium: $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$ ...(i)
So one mole of $MnO_4^-$ accepts 5 moles of electrons.
Ferrous oxalate is $FeC_2O_4$; it contains both $Fe^{2+}$ and $C_2O_4^{2-}$, and both are oxidised by permanganate.
$Fe^{2+} \rightarrow Fe^{3+} + e^-$ (1 electron)
$C_2O_4^{2-} \rightarrow 2CO_2 + 2e^-$ (2 electrons) ...(ii)
Total electrons lost by one mole of $FeC_2O_4$ $= 1 + 2 = 3$ moles.
Moles of $MnO_4^-$ required $= \dfrac{3}{5} = 0.6$ mole
Balanced equation: $3MnO_4^- + 5FeC_2O_4 + 24H^+ \rightarrow 3Mn^{2+} + 5Fe^{3+} + 10CO_2 + 12H_2O$
Note: the source solution considers only the oxalate part: $2MnO_4^- + 5C_2O_4^{2-} + 16H^+ \rightarrow 2Mn^{2+} + 10CO_2 + 8H_2O$, i.e. 2 moles of $MnO_4^-$ for 5 moles of oxalate, which gives $\dfrac{2}{5} = 0.4$ mole for one mole of oxalate. For complete oxidation of ferrous oxalate the $Fe^{2+}$ must also be counted, which gives 0.6 mole.