Let $H$ be a complex Hilbert space and $S,T \in B(H)$ partial isometries. Then $S T$ is a partial isometrie, if and only if $T^*(\ker(S)) \subseteq \ker(ST)$.
Edit: My attempts so far:
$\Rightarrow$: Let $x \in \ker(S)$, then $||STT^*(x)||^2=<STT^*(x), STT^*(x)>=<PT^*(x),T^*(x)>$, where P is the orthogonal projection onto $ker(ST)^\perp$. I've tried many different angles but this is where I always end up
Let's introduce the projections $P=S^*S$ (onto $(\ker S)^\perp$), and $Q=TT^*$ (onto $\operatorname{ran}T$). The condition of $ST$ being partial isometry is $ST=ST(ST)^*ST$ which simplifies to $ST=SQPT$. Rewrite this further as $S(I-QP)T=0$. Since $\ker S=\ker P$ and $\operatorname{ran} T = \operatorname{ran} Q$, the latter is equivalent to $P(I-QP)Q =0$. Thus, $ST$ is a partial isometry if and only if $$PQ = PQPQ \tag1$$
Next, look at the condition $T^*(\ker(S)) \subseteq \ker(ST)$. It is equivalent to $TT^*(\ker(S)) \subseteq \ker(S )$. Rewrite it as $Q(\ker P) \subseteq \ker P$. Express this as $PQ(I-P)=0$. So, the condition is equivalent to $$PQ = PQP \tag2$$
Now it should be clear that (2) implies (1), and you need to work on the converse.