The limits are always the same.

As $\mathcal K = S_2(H)$ is a Hilbert space, and as only the Schur product is involved anywhere, we can infact identify $\mathcal K$ with $\ell^2 = \ell^2(\mathbb N)$ and consider the pointwise product of vectors in $\ell^2$.

Let $x=(x_r)\in\ell^2$ and let $(y^{(i)})$ be a bounded sequence in $\ell^2$. Thus, for each $r$ the scalar sequence $(y^{(i)}_r)$ is bounded, so by moving to subsequences and using a diagonal argument, we may suppose that $y^{(i)}_r\rightarrow y_r$, say, as $i\rightarrow\infty$. Set $y=(y_r)$ and observe that for any $R$,
$$ \sum_{r=1}^R |y_r|^2 = \sum_{r=1}^R \lim_i |y^{(i)}_r|^2
= \lim_i \sum_{r=1}^R |y^{(i)}_r|^2
\leq \lim_i \|y^{(i)}\|_2^2 < \infty. $$
As $R$ was arbitrary, we conclude that $y\in\ell^2$.

Then, for $R>0$,
\begin{align*} \lim_i \|xy^{(i)} - xy\|_2^2
&= \lim_i \sum_r |x_r y^{(i)}_r - x_r y_r|^2 \\\\
&= \sum_{r=1}^R \lim_i |x_r y^{(i)}_r - x_r y_r|^2
+ \lim_i \sum_{r>R} |x_r y^{(i)}_r - x_r y_r|^2 \\\\
&= \lim_i \sum_{r>R} |x_r y^{(i)}_r - x_r y_r|^2 \\\\
&\leq \lim_i \Big(\sum_{r>R} |y^{(i)}_r - y_r|^2\Big) \Big( \sup_{r>R} |x_r| \Big).
\end{align*}
We can make this arbitrarily small by choosing $R$ large. We conclude that $xy^{(i)} \rightarrow xy$ in norm.

Given bounded sequences $(x^{(i)}), (y^{(i)})$ in $\ell^2$, let $x,y$ be the pointwise limits, as in the previous paragraph. Let $z\in\ell^2$, and consider
\begin{align*} \lim_i \sum_r x^{(i)}_r z_r y^{(i)}_r. \end{align*}
As $x^{(i)} y^{(i)}$ is in $\ell^1$ and $(z_r)$ is bounded, this sum make sense, and we can copy the argument to show that
\begin{align*} \lim_i \sum_r x^{(i)}_r z_r y^{(i)}_r
= \sum_r x_r z_r y_r
. \end{align*}

Thus, given bounded sequences $(x^{(i)}), (y^{(i)}), (a^{(i)}), (b^{(i)})$ in $ell^2$, let $x,y,a,b$ be the pointwise limits, as in the previous paragraph. Given a bounded linear map $\phi$ on $\ell^2$, we see that as $a^{(i)} b^{(j)} \rightarrow a b^{(j)}$ in norm, as $i\rightarrow\infty$, also $\phi(a^{(i)} b^{(j)}) \rightarrow \phi(a b^{(j)})$ in norm. Hence, by the same argument,
\begin{align*}
\lim_j \lim_i \sum_r x^{(i)}_r y^{(j)}_r \phi(a^{(i)} b^{(j)})_r
= \lim_j \sum_r x_r y^{(j)}_r \phi(a b^{(j)})_r
= \sum_r x_r y_r \phi(a b)_r.
\end{align*}
By symmetry, we get the same answer with the limits taken in the other order.