Let $T\in\mathcal{B}(\mathcal{H})$ be an operator on a Hilbert space, and let $T = V|T|$, $T^* = W|T^*|$ be the unique polar decompositions such that $\text{ker}(V) = \text{ker}(T)$ and $\text{ker}(W) = \text{ker}(T^*)$ respectively. One can check that $VW$ is itself a partial isometry: $$ (VW)(VW)^*(VW) = VWW^*V^*VW = VP_{\text{ran}W}P_{\text{ker}V^\perp}W $$ but $\text{ker}(V)^\perp = \text{ker}(T)^\perp = \overline{\text{ran}(T^*)}\subseteq\text{ran}(W)$ (partial isometries have closed ranges), so $P_{\text{ran}W}P_{\text{ker}V^\perp} = P_{\text{ker}V^\perp}$ and $VP_{\text{ker}V^\perp} = V$.
What are the initial and final spaces of $VW$? Clearly $\text{ker}W\subseteq\text{ker}VW$, but in general we need to know when $Wh\in\text{ker}V$, and I don't see any easy way of working with $\text{ran}W\cap\text{ker}V$. Is there anything nice we can do here?
Edit: In Kadison & Ringrose's Fundamentals of Operator Theory, Theorem 6.12, they claim that the same partial isometry $V$ gives us $T = V|T| = |T^*|V$, which would imply $T^* = V^*|T^*|$. They also claim $\text{ran}(V) = \overline{\text{ran}(T)}$ (well, they say $=\text{ran}(T)$, but this can't possibly be true since the range of $T$ isn't always closed whereas $\text{ran}(V)$ is closed, so I take that as a typo). This implies $\text{ker}(V^*) = \text{ran}(V)^\perp = (\overline{\text{ran}(T)})^\perp = \text{ker}(T^*)$, and by uniqueness of the polar decomposition this would imply $W = V^*$. Is this true? It's certainly the case for the shift operator on $\ell^2(\mathbb{N})$ (the only operator I bothered calculating this for).
$\def\ran{\operatorname{ran}}$ $\def\abajo{\\[0.3cm]}$The uniqueness in the polar decomposition gives you a bit more. Namely, $$ V^*V=P_{\overline\ran T^*},\qquad\qquad VV^*=P_{\overline\ran T} $$ and $$ W^*W=P_{\overline\ran T},\qquad\qquad WW^*=P_{\overline\ran T^*}. $$ So $$ V^*V=WW^*,\qquad\qquad VV^*=W^*W.$$ Then $$ VW(VW)^*=VWW^*V=VV^*VV^*=VV^*=P_{\overline\ran T}, $$ and $$ (VW)^*VW=W^*V^*VW=W^*WW^*W=W^*W=P_{\overline\ran T}. $$ So the initial and final projection of $VW$ is the range projection for $T$.
With a bit more subtlety, we can do more. We have $$ T^*(TT^*)^n=(T^*T)^nT^*. $$ Taking linear combinations and limits, $T^*f(TT^*)=f(T^*T)T^*$ for all continuous functions $f$. In particular $$ T^*|T^*|=|T|\,T^*. $$ We know that $\ker W=\ker T^*$. If $z\in (\ker W)^\perp=(\ker T^*)^\perp=\overline\ran T$, from $\overline\ran T=\overline\ran TT^*$ we can write $z=TT^*y$ for some $y$. Then \begin{align} VWz&=VWTT^*y=VW|T^*|^2y=VW|T^*|\,|T^*|y\abajo &=VT^*|T^*|y=V|T|\,T^*y=TT^*y=z. \end{align} That is, $$ VW=P_{\overline\ran T}=VV^*. $$ Applying $V^*$ on the left, and using that $V^*V=P_{\overline\ran T}=P_{\overline\ran V^*}=P_{\overline\ran W}$, we get $$ W=V^*. $$