Two cuts. Label the beads modulo $2(k+m)$ as $0,1,2, \cdots, 2(k+m)-1$. Then consider intervals of length $k+m$, of the form $[x,x+(k+m)-1]$. As $x$ changes by one, the number of white beads changed by at most one. If there are $\ell$ white beads in this interval when $x=0$, when $x=k+m$ there will be $2k-\ell$ white beads. Because $\min(\ell, 2k-\ell) \le k \le \max(\ell, 2k-\ell)$ by the discrete intermediate value theorem there is some $x$ so that this interval has $k$ white beads. As it has $k+m$ total beads, it has $m$ black beads. Similar holds true for the complement, so we can use two cuts. Clearly one cut doesn't work, because the chain will still be in one piece. For a far-reaching generalization, see https://www.tau.ac.il/~nogaa/PDFS/Publications2/Splitting%20necklaces.pdf and https://www.youtube.com/watch?v=rwiEiGqgetU.