In this answer, I will attempt to explain using chemical principles, rather than detailed mathematics.
As has been said already, reactions 1 and 3 both consume $\ce{S2O8^{2-}}$. Now, for the rate determining step to dominate the decay of this species, there would necessarily be very little consumption of it in any other steps. IE, there would be little consumption of it in reaction 3, implying that there is very little supply of the $\ce{CO2^-}$ radical.
Now, unfortunately for the complexity of the kinetics, there is a rapid interconversion of $\ce{SO4^-}$ to $\ce{CO2^-}$ by reaction with formate, which in turn rapidly reacts with more $\ce{S2O8^{2-}}$ to reform the $\ce{SO4^-}$ which can react again with formate, you get the picture. So, although there will only be a very small concentration of $\ce{SO4^-}$ and $\ce{CO2^-}$ present (by virtue of the slow reaction in step 1), the sheer relative speed of formate induced interconversion makes these other steps important. Of course, the more formate we have, the more $\ce{CO2^-}$ we can liberate (to a point, hence the square root), and the faster we can remove $\ce{S2O8^{2-}}$. This is, in fact, a catalytic cycle within the greater reaction scheme.
This is what is meant by the steady state approximation. As the concentrations of $\ce{SO4^-}$ and $\ce{CO2^-}$ are always very small, we can assume that their total rate of change across all reactions in the scheme is (extremely close to) zero. This allows us to write a set of simultaneous equations from which we can try to determine the (often quite complex) rate law by estimating the "steady state concentrations" of the low concentration species in terms of the other species of interest.
