It seems OP understands Poutnik's rationalization on the reduction step, so this question does not need an answer. However, I feel it is not completed because OP's statement of:
[...] but as the oxidation number of sulphur in $\ce{S2O8^2−}$ and $\ce{SO4^2−}$ is the same, i.e., equal to +6, [...]
This statement is not quite true if you just do the calculations and conclude from that as OP initially did. For benefit of other novice chemistry readers, I think it is better to explain what's going on in this situation in detail.
Suppose the oxidation number of sulfur in peroxodisulphate ion $(\ce{S2O8^2−})$ is $\alpha$. Then if you do usual calculations, it is like:
$$ 2 \cdot \alpha + 8(-2) = -2 \ \Rightarrow \ \therefore \alpha = \frac{16 - 2}{2} = +7$$
This means there is a reduction happening. However, this calculation is also erroneous. Let's look at the Lewis structure of the peroxodisulphate ion, which is a particularly interesting one:
The trick here is that you should be able to notice that $\ce{S2O8^2−}$ ion has two oxygen atoms that are bonded via a single bond, which is known as a peroxide linkage. This linkage will influence the average oxidation state of oxygen in $\ce{S2O8^2−}$ ion. The two oxygen atoms in peroxide linkage will have a $−1$ oxidation state, similar to the same oxidation state that oxygen has in all peroxides. The rest of oxygens in the anion have usual $−2$ oxidation state, which implies that the average oxidation state of oxygen in $\ce{S2O8^2−}$ ion is:
$$\frac{6(-2) + 2(-1)}{8} = -\frac74$$
Thus, if we redo the calculations:
$$ 2 \cdot \alpha + 8(-\frac74) = -2 \ \Rightarrow \ \therefore \alpha = \frac{14 -2}{2} = +6$$
This did not imply a reduction when we look at the oxidation number of sulfur in $\ce{SO4^2−}$ ion. If it is $\beta$ in $\ce{SO4^2−}$, then,
$$ \beta + 4(-2) = -2 \ \Rightarrow \ \therefore \beta = 8 - 2 = +6$$
Yet, if you know the Lewis structure of the peroxodisulphate ion, you would realize there is something going on for oxygen atoms in $\ce{S2O8^2−}$ ion, which cannot be explained by these calculations. Therefore, it is always better to conclude by the facts found using the redox half reactions. Consequently, I admit that I also prefer to do usual redox half reactions if I find no change in oxidation states doing usual calculations, because that gives no place to make erroneous calculations. Accordingly, for the reduction half reaction:
$$\ce{S2O8^2− + 2 e- <=> 2SO4^2- } \tag1$$
This is the reduction half reaction. The equation $(1)$ clearly indicates the $\ce{S2O8^2−}$ ion has undergone a reduction during the complete reaction. The corresponding oxidation half reaction would be:
$$\ce{Mn^2+ + 4H2O <=> MnO4- + 8H+ + 5 e-} \tag2$$
If you combine the equations $(1)$ and $(2)$ in order to cancel the electrons, you get:
$$\ce{2Mn^2+ + 8H2O + 5 S2O8^2- -> 2MnO4- + 16H+ + 10 SO4^2-} \tag3$$
This is the given equation. When you consider the Standard Electrode (Reduction) Potentials $(E^\circ)$ in aqueous solutions at $\pu{25 ^\circ C}$, you would find for reaction $(1)$ it is $\pu{2.01 V}$ and for reaction $(2)$ it is $\pu{-1.51 V}$ (oxidation potential). Thus, for the completed reaction $(3)$ at standard conditions, $E^\circ_\mathrm{Rxn} = 2.01 + (-1.51) = \pu{+0.50 V}$, implying it is spontaneous at standard conditions.
