How to assign oxidation number of sulfur in thiocyanate anion?
In the end the oxidation state of sulfur does not really matter for balancing equations.  Rather, identifying the real oxidizing and reducing agents is.

Clearly the iodine in the iodate ion is the oxidizing agent, dropping from $+5$ to $+1$.  But copper, sulfur, and even carbon are all being oxidized or reduced from the cuprous thiocyanate.   We have multiple parts of the $\ce{CuSCN}$ participating in the electron exchange.

Why not then identify the whole $\ce{CuSCN}$ molecule as the reducing agent, thus covering all the reducing agent atoms in their proper proportions?  With this approach you see that the sum of oxidation states for this reductant goes from $0$ to $+7$ ($+2$ for $\ce{Cu}$ in the copper sulfate, $+6$ for the sulfur in the same compound, and $-1$ for $\ce{CN}$ in the hydrogen cyanide).  Then $\ce{CuSCN}$ gives $7$ electrons, iodate takes $4$ electrons, and thus you get $\ce{4 CuSCN + 7 KIO3}$, just like with the matrix method.