Atomic radii of Sc, Ti, Fe, and Co
There are different notions of atomic radius; the one you're using seems to be the metallic radius, which is half the distance between nearest neighbors in the metal.  This notion is very sensitive to the number of electrons per atom involved in bonding.  Scandium has only 3 valence electrons, while $\ce{Ti}$ has 4.  These all participate, in some extent, in the "electron soup" that holds metals together.  I haven't been able to figure out exactly to what extent, but  It's fair to say though that the 4 valence electrons of Ti bind the nuclei together significantly more tightly than the 3 of $\ce{Sc}$. As a result, the $\ce{Ti}$ atoms get significantly closer together.  (An analogous situation is the covalent radius of $\ce{F2}$, of around $\pu{70 pm}$, versus that of $\ce{O2}$, of around $\pu{60 pm}$; although the covalent radius tends to decrease across a period, it grows from $\ce{O}$ to $\ce{F}$ because $\ce{F2}$ has a single bond while $\ce{O2}$ has a double one.) 

As you progress further along the transition metals the delocalization of d electrons in the metal goes down.  That is, although there are more d electrons in $\ce{Co}$ than in $\ce{Fe}$, their effectiveness in binding atoms together isn't really any larger.  As a result, the distance between neighbors (hence the metallic radius) is the same for both.