Given a von Neumann algebra $M$ on an Hilbert space $H$, my naive understanding is that if I have a partial isometry $v\in M$ and projection $p\in P(M)$ such that $v^* v = p$, where $v^*$ is the adjoint, then the range of $v$ and its kernel are not necessary disjoint, because I could always change the $v$ by multiplying it with a unitary, that maps some part of the range of $v$ into its kernel, so I would not exclude the possibility of an $0 \neq \xi \notin Kernel(v)$ such that $vv \xi = 0$. However, if p is a central projection it seems to me that this is not true from the following argument:
Assume $\xi \neq 0$ and $\xi \notin Kernel(v)$ but $v^2 \xi = 0$. Then $p \xi = p^2 \xi =p v^* v \xi = v^* p v \xi = v^* v^* v v \xi = 0$, where I used that $p$ is central in the third equality. Thus $\xi \in Kernel(p)$. But since $Kernel(p) = Kernel(v)$ we have $\xi \in Kernel(v)$ which is a contradiction.
Does this work?
I think your argument is fine. Another way to write your idea is to note that the equality $$p=v^*v=v^*vp=v^*pv=(v^*)^2v^2$$ implies that $\ker v^2=\ker p=\ker v$. This shows that $\def\ran{\operatorname{ran}}$ $\ran v\cap \ker v=\{0\}$.