I don't think so, even in the case of $M=S^1$ and $k=2$: $F(M,2)$ would be homotopy equivalent to $S^1$ and $F(W(M),2)$ would be $S^1\times S^1$ component glued to $S^1\times$ ribbons...

Here's some handwaving on the above: for $F(S^1,2)$ pick the 1st point on the circle, and the 2nd point can be only on $S^1$ \\ $pt$, which makes $F(S^1,2)$ an open interval fibration over $S^1$, thus homotopy equivalent to $S^1$.

On the other hand, when you pick the 1st point on $W(S^1)$ there are 3 possibilities: you pick it on the gluing point, you pick it on the circle but not on the gluing point, or you pick it on the open interval connected to the circle via the gluing point. Wait a sec, I have doubts now: have to figure out how exactly those parts are glued before claiming an $S^1\times S^1$ component...