$\newcommand\range{\operatorname{range}}
\newcommand\B{\mathcal B}
\newcommand\H{\mathcal H}
\newcommand\K{\mathcal K}
$
We can achieve the universal property for certain bilinear maps of the specific form $F(x,y) = H(x) K(y)$. More precisely:

**Theorem:** Let $\mathcal H,\mathcal K,\mathcal L$ be Hilbert spaces. Let $H:\mathcal B(\mathcal H)\to\mathcal B(\mathcal L)$ and $K:\mathcal B(\mathcal K)\to\mathcal B(\mathcal L)$ be normal unital $*$-homomorphisms. Assume that the ranges of $H$ and $K$ commute. Then there is a normal unital *-homomorphism $G:\mathcal B(\mathcal H\otimes\mathcal K)\to \mathcal B(\mathcal L)$ such that $G(h\otimes k)=H(h)K(k)$.

**Proof:** By this answer, $H(h)=U_H(h\otimes 1_{\mathcal H_0})U_H^*$ for some Hilbert space $\mathcal H_0$ and some unitary $U_H$. Thus $H=\hat H \circ \iota_H$ where $\hat H(x):=U_HxU_H^*$ and $\iota_H(h)=h\otimes 1_{\mathcal H_0}$. $\hat H$ is a unital $*$-isomorphism and therefore normal.
$\iota_H$ is a unital $*$-homomorphism. $\iota_H$ is weak*-continuous and therefore normal.
Analogously, we get that $K=\hat K\circ\iota_K$ for some normal unital *-isomorphism $\hat K$ and the normal unital *-homomorphism $\iota_K(k):=k\otimes 1_{\mathcal K_0}$. Let $\iota_0(x):=1_{\mathcal H}\otimes x$ for $x\in\mathcal H_0$, also a normal unital *-homomorphism.

The ranges of $H,K$ commute by assumption. Thus $\hat H(\mathcal B(\mathcal H)\otimes \mathbf 1)$ and $\hat K(\mathcal B(\mathcal K)\otimes \mathbf 1)$ commute. Here $\mathbf 1$ denotes the von Neumann algebra in $\mathcal B(\mathcal H_0),\mathcal B(\mathcal K_0)$, respectively, that consists of the multiples of identity. And $\otimes$ is tensor product of von Neumann algebras. Since $\hat H$ is a normal *-isomorphism, $\mathcal B(\mathcal H)\otimes\mathbf 1$ and $S:=\hat H^{-1}\hat K(\mathcal B(\mathcal K)\otimes\mathbf 1)$ commute and $S$ is a von Neumann algebra. Hence $$S \subseteq (\mathcal B(\mathcal H)\otimes\mathbf 1)' \stackrel{(*)}= \mathcal B(\mathcal H)'\otimes \mathbf 1' = \mathbf 1\otimes \mathcal B(\mathcal H_0) = \range \iota_0.$$
Here $X'$ denotes the commutant of $X$ and $(*)$ follows from the commutation theorem for tensor products of von Neumann algebras.

Since $\iota_0$ is an isometry, $\iota_0^{-1}:\mathbf 1\otimes\mathcal B(\mathcal H_0)\to\mathcal B(\mathcal H_0)$ exists and is a normal unital $*$-isomorphism.
Thus
$$
f : \mathcal B(\mathcal K)
\overset{\iota_K}\longrightarrow
\mathcal B(\mathcal K) \otimes \mathbf 1
\overset{\hat H^{-1}\hat K}\longrightarrow
S
\overset{\iota_0^{-1}}\longrightarrow
\mathcal B(\mathcal H_0)
$$
is a composition of unital normal *-homomorphisms and hence a unital normal *-homomorphism.

Since $1_{\B(\H)}$ and $f$ are normal $*$-homomorphisms and thus completely positive, by [Tak, Prop. 5.13], $1_{\B(\H)}\otimes f:\B(\H\otimes\K)\to\B(\H\otimes\H_0)$ exists and is normal and completely positive. And it satisfies $(1_{\B(\H)}\otimes f)(h\otimes k)=1(h)\otimes f(k)$.
Then $1_{\B(\H)}\otimes f$ is also unital and bounded.
On the algebraic tensor product $\B(\K)\otimes_{\mathit{alg}}\B(\H)$, $1_{\B(\H)}\otimes f$ is a
*-homomorphism, and thus it is a *-homomorpism everywhere by continuity.
Let $G := \hat H\circ (1_{\mathcal B(\mathcal H)}\otimes f)$.
Since $\hat H$ is also a normal unital $*$-homomorpishm, so is $G$.

We have
$$
G(h\otimes 1_{\mathcal K}) = \hat H(h\otimes 1) = H(h)
$$
and
$$
G(1_{\mathcal H}\otimes k) =
\hat H(1_{\mathcal H}\otimes f(k)) =
\hat H(\iota_0\circ f(k)) =
\hat H(\hat H^{-1}\circ\hat K\circ \iota_K(k)) =
\hat K\circ \iota_K(k) = K(k).
$$
Since $G$ is a *-homomorphism, this implies $G (h\otimes k) = G(h\otimes 1)G(1\otimes k)=H(h)K(k)$ as desired.

[Tak] *Takesaki, M.*, Theory of operator algebras I., Encyclopaedia of Mathematical Sciences 124. Operator Algebras and Non-Commutative Geometry 5. Berlin: Springer (ISBN 3-540-42248-X/hbk). xix, 415 p. (2002). ZBL0990.46034.