Q:

Is the sum of separating vectors always separating?

If $\mathcal{R}$ is a von Neumann algebra acting on Hilbert space $H$ and $v, w \in H$ are separating vectors for $\mathcal{R}$, must $v+w$ be (either zero or) separating for $\mathcal{R}$?
[I have edited to remove the restriction to type III factors and am moving my proposed partial solution to an answer below.]

A:

No, there must be a counterexample, under the mild assumption that there exists a nontrivial unitary $U \in \mathcal{R}$ whose restriction to the range of some nonzero projection $P \in \mathcal{R}$ is trivial (i.e. the identity).
Fix such a $U$ and $P$. Let $v$ be any separating vector for $\mathcal{R}$ and let $w = -Uv$. This $w$ is separating for $\mathcal{R}$ since any nonzero $T \in \mathcal{R}$ that annihilated $w$ would make $-TU$ a nonzero operator in $\mathcal{R}$ than annihilates $v$.
But we can show, using the fact that $UP = P$ and $U(1-P) = (1-P)U$, that $v + w$ is not separating for $\mathcal{R}$:
$v + w = v - Uv = (Pv + (1-P)v) - (UPv + U(1-P)v)$
$= (1-P)v - U(1-P)v = (1-P)v - (1-P)Uv = (1-P)(1-U)v$;
and $(1-P)(1-U)v$ is annihilated by $P$.

