Let $H_1$ and $H_2$ be complex vector spaces equipped with non-negative sesqui-linear forms $\langle \cdot ,\cdot \rangle _1$, and $\langle \cdot ,\cdot \rangle _2$, respectively, so that each can be viewed as a pre-Hilbert space (with possibly degenerate norms).
To a large extent you are welcome to assume that $H_1$ and $H_2$ are indeed Hilbert spaces.
Also let $$ \varphi :H_1\times H_2\to \mathbb C \qquad \qquad (*) $$ be a sesqui-linear form such that $$ |\varphi (x_1,x_2)|^2\leq \langle x_1,x_1\rangle _1\langle x_2,x_2\rangle _2, \quad\forall (x_1,x_2)\in H_1\times H_2. $$ Should $H_1$ and $H_2$ be Hilbert spaces, this would be simply written as $$ |\varphi (x_1,x_2)|\leq \|x_1\|\|x_2\|. $$
For $x=(x_1, x_2)$ and $y=(y_1, y_2)$ in $H_1\times H_2$, put $$ \langle x,y\rangle _\varphi = \langle x_1,y_1\rangle _1 +\varphi (x_1, y_2)+ \overline {\varphi (y_1, x_2)}+ \langle x_2,y_2\rangle _2. $$
Warm up exercise 1. Prove that, for every $x=(x_1, x_2)\in H_1\times H_2$, one has that $\langle x,x\rangle _\varphi \geq 0$.
We may then give the following:
Definition. The semi-direct product of $H_1$ and $H_2$ relative to $\varphi $, denoted $H_1\bowtie_\varphi H_2$, is the Hilbert space completion of $H_1\times H_2$ under the pre-inner product $\langle \cdot ,\cdot \rangle _\varphi $, after the subspace formed by the vectors of norm zero is modded out.
Observe that the map $$ \iota _1:x_1\in H_1\mapsto (x_1, 0)\in H_1\bowtie_\varphi H_2 $$ may be viewed as an isometry in the sense that $$ \|\iota _1(x_1)\| = \sqrt{\langle x_1,x_1\rangle_1 }, \quad\forall x_1\in H_1, $$ and similarly for $H_2$.
It is then easy to show that $$ \langle\iota_1(x_1),\iota_2(x_2)\rangle=\varphi(x_1,x_2), $$ for all $x_1\in H_1$ and $x_2\in H_2$.
Warm up exercise 2. Given any Hilbert-space $H$, and given subspaces $H_1, H_2\subseteq H$, consider the sesqui-linear form $$ \varphi :(x_1, x_2)\in H_1\times H_2\to \langle x_1, x_2\rangle \in \mathbb C, $$ and prove that there is an isometric linear map from $H_1\bowtie_\varphi H_2$ to $H$, sending each $(x_1, x_2)\in H_1\times H_2$ to $x_1+x_2$.
Warm up super trivial exercise 3. Prove that $H_1\bowtie_0 H_2\simeq H_1\oplus H_2$.
Ok, now to the real stuff:
Question. Find natural conditions on the sesqui-linear form $\varphi $ given in $(*)$, above, such that the orthogonal projections from $H_1\bowtie_\varphi H_2$ to $H_1$ and to $H_2$ commute with each other.
PS 1: The motivation for this question is to outline a possible approach for answering Do GNS-projections of different tensor factors commute?, in the special case of pure states.
The context there is of a pair of $C^*$-algebras $A$ and $B$ (which I'm assuming unital for simplicity), and a state $\omega $ on the (maximal) tensor product $A\otimes B$. One can then define sesqui-linear forms on $A$ and $B$ by $$ \langle x,y\rangle _A = \omega \big ((y^*x)\otimes 1\big ),\quad\forall x,y\in A, $$ and $$ \langle x,y\rangle _B = \omega \big (1\otimes (y^*x)\big ),\quad\forall x,y\in B, $$ and further define $$ \varphi :A\times B\to {\mathbb C}, $$ by $$ \varphi (x,y)=\omega (x\otimes y^*). $$ In case $\omega $ is a pure state, one would then attempt to check whether or not $\varphi $ obeys the conditions arising from the answer to the above question.
PS 2: Another natural question would be to try to determine conditions on the sesqui-linear form $\varphi $ given in $(*)$, above, such that $H_1+H_2$ is closed in $H_1\bowtie_\varphi H_2$.
Edit: PS 3: Should $H_1$ and $H_2$ be Hilbert spaces, then by Riesz there exists a bounded operator $T:H_1\to H_2$, such that $$ \varphi(x_1,x_2) =\langle T(x_1),x_2\rangle_2. $$ In this case, an answer to the above question is: the projections mentioned there commute iff $T$ is a partial isometry, that is, $TT^*T=T$.
Although this is a nice result, I wouldn't consider it as being a condition on the sesqui-linear form $\varphi$ since it relies too much on geometric aspects of operators. In particular this would be hard to check in the application to (PS1).