Yes, an operator is compact if it maps the unit ball into a precompact set. This is the usual definition for Banach spaces. For Hilbert spaces the same applies (see e.g. these notes). Indeed, a bounded linear $T:H_1\rightarrow H_2$ is compact if and only if $T^*T$ is compact, iff $TT^*$ is compact, iff $T^*$ is compact.
To deal with the second question, I think one proceeds as follows. Firstly, observe that there is a $*$-algebra isomorphism
$$ \mathcal B(H_1\oplus H_2) \cong \begin{pmatrix}
\mathcal B(H_1) & \mathcal B(H_2, H_1) \\
\mathcal B(H_1, H_2) & \mathcal B(H_2)
\end{pmatrix}. $$
To see this, think about the matrix acting on $H_1\oplus H_2$ written as a column vector. Under this isomorphism, compact operators behave as you might hope,
$$ \mathcal K(H_1\oplus H_2) \cong \begin{pmatrix}
\mathcal K(H_1) & \mathcal K(H_2, H_1) \\
\mathcal K(H_2, H_1) & \mathcal K(H_2)
\end{pmatrix}. $$
Thus I can speak of $\mathcal K(H_1,H_2)$ as a "corner" of $\mathcal K(H_1\oplus H_2)$.
So, one can define $\mathcal K(H_1,H_2) \otimes A$ as the closure of $\mathcal K(H_1,H_2) \odot A$ inside $\mathcal K(H_1\oplus H_2) \otimes A$. Things are nicer than this: let $p_i$ be the projection of $H_1\oplus H_2$ onto $H_i$. Then $p_2 \mathcal K(H_1\oplus H_2) p_1$ is isomorphic to $\mathcal K(H_1,H_2)$, and $p_i\otimes 1\in M(\mathcal K(H_1\oplus H_2)\otimes A)$. One can check that $\mathcal K(H_1,H_2) \otimes A$ is isomorphic to $(p_2\otimes 1)(\mathcal K(H_1\oplus H_2)\otimes A)(p_1\otimes 1)$.
The notes by Maes and Van Daele are nice, but I do find that there are various little inaccuracies, or points like this which are not (well) explained. The original papers by Woronowicz are terse, but I think a pleasure to read, and will improve your intuition about the subject. You could also look at the book of Timmermanns, but that takes a different approach.
In a comment, Ruy points our how to get this tensor product via representing on a Hilbert space.
An approach using more theory would be to use Hilbert $C^\ast$-modules (I follow chapter 4 of Lance's book). $\mathcal K(H_1, H_2)$ is a right module over $\mathcal K(H_1)$ for the "inner-product" $(S|T) = S^*T$. $A$ is a module over itself. The exterior tensor product of modules gives $\mathcal K(H_1, H_2) \otimes A$ as a right module over $\mathcal K(H_1)\otimes A$. A rather tedious check shows that the norms of all three approaches are the same.
For the application, we need:
- For a state $h\in A^*$ we want to make sense of $\iota\otimes h$ as a map $\mathcal K(H_1,H_2)\otimes A \rightarrow \mathcal K(H_1,H_2)$;
- For a $*$-homomorphism $\Phi:A\rightarrow A\otimes A$ we need to make sense of $\iota\otimes\Phi$ as a homomorphism (suitable interpreted) $\mathcal K(H_1,H_2)\otimes A \rightarrow \mathcal K(H_1,H_2)\otimes A \otimes A$.
I think my original approach, of viewing things as a "corner" of $C^\ast$-algebra, is probably the easiest way to get these properties.
