Let $M$ be a von Neumann algebra in $B(H)$ and assume $\zeta$ is a cyclic vector for $M$. There is a deep result which says: for a given vector $\eta$ in $H$, there exists an operator $x\in M$ with $\zeta=x\eta$.
Question. Assume $\zeta$ and $\eta$ have the same norm. Can we say that there exists a partial isometry $u\in M$ with $\zeta=u\eta$?
As stated, the result you quote isn't true: for instance, let $M=L^\infty [0,1 ]\subset B (L^2 [0,1]) $ acting by multiplication. Then $M $ has a cyclic vector (any constant function, for instance) but if $\eta $ is bounded and $\zeta $ is unbounded, no $x $ with $\zeta=x\eta$ can exist.
Kadison's Transitivity Theorem also has the requirement that $M $ (not necessarily a von Neumann algebra) is irreducible. So, in your case, $M=B (H) $ and the answer is trivially yes. The power of the theorem comes when applied to an irreducible non-von-Neumann C$^*$-algebra; but in that case the answer to your question would be negative, since a C$^*$-algebra may lack partial isometries at all.