= Solution
A closed subspace invariant under a <star-algebra> is reducing, since it is invariant under every adjoint as well. Thus the <orthogonal projections> $P_1,P_2$ onto $H_1,H_2$ commute with $M$. Consider the bounded intertwiner
$$
T=(I-P_2)|_{H_1}:H_1\longrightarrow H_2^\perp.
$$
Its <kernel> is $H_1\cap H_2$, so its initial polar space is
$$
K_1=H_1\cap(H_1\cap H_2)^\perp.
$$
Its closed range is
$$
K_2=\overline{H_1+H_2}\cap H_2^\perp.
$$
For the inclusion from left to right, $(I-P_2)h_1=h_1-P_2h_1$ lies in $H_1+H_2$ and is orthogonal to $H_2$. Conversely, approximate any $z$ in the displayed $K_2$ by $h_{1,n}+h_{2,n}$ and apply $I-P_2$. The images converge to $z$ and equal $Th_{1,n}$, proving the reverse inclusion after <closure>. The overline on the sum is present in the original PDF and is essential for potentially nonclosed sums.
Take the polar decomposition $T=V|T|$ between these <Hilbert spaces>. Because $T$ intertwines the two actions of $M$, its adjoint does also, so $T^*T$ commutes with the action on $H_1$. Its positive square root therefore commutes as well. The identity $V|T|\xi=T\xi$ shows on its dense initial range that $V$ intertwines, and <continuity> extends this to $K_1$. It is an isometry onto $K_2$. Consequently
$$
\boxed{H_1\cap(H_1\cap H_2)^\perp
\ \cong_M\ \overline{H_1+H_2}\cap H_2^\perp.}
$$
The unitary here is a unitary between the indicated <Hilbert space modules>, not an assertion that the unclosed algebraic sum is a <Hilbert space>.
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