Notice that
$$
|AT|^2 = T^*A^*AT \leq \|A\|^2T^*T = \|A\|^2|T|^2,
$$
so, by Proposition 1.3.8 in [1] (square-root is an operator monotone function), one has that
$$
|AT| \leq \|A\||T|,
$$
and it follows that $AT$ is trace class according to the OP's definition.
[1] Pedersen, Gert K., C*-algebras and their automorphism groups, London Mathematical Society Monographs. 14. London - New York -San Francisco: Academic Press. X, 416 p. $ 60.00 (1979). ZBL0416.46043.
answer 1: I am not exactly sure, what you mean by the "original (algebraic)" tensor product.
Usually, for two Hilbert spaces $ H_1, H_2 $ the algebraic tensor product $ H_1 \otimes_{alg} H_2 $ is the space of all linear combinations of vectors $ \phi_1 \otimes \phi_2 $. As explained in the Wikipedia article, you define an inner product $ \langle \cdot , \cdot \rangle $ on it by
$$\langle \phi_1 \otimes \phi_2, \psi_1 \otimes \psi_2 \rangle := \langle \phi_1 , \psi_1 \rangle_1 \; \langle \phi_2, \psi_2 \rangle_2.$$
Now, $ H_1 \otimes_{alg} H_2 $ with this inner product is not a Hilbert space, since it is not complete. But its completion $ H_1 \otimes H_2 := \overline{H_1 \otimes_{alg} H_2}^{\langle \cdot , \cdot \rangle} $ is a Hilbert space and is called the tensor product Hilbert space.
The space of Hilbert-Schmidt operators $ HS(H_1^*,H_2) $ you mentioned above is isometrically isomorphic to $ H_1 \otimes H_2 $. You may identify all tensor products $ \phi_1 \otimes \phi_2 $ with the map $ x^* \mapsto x^*(\phi_1) \phi_2 $. This map is in $ HS(H_1^*,H_2) $. Taking linear combinations and limits, this identification extends to an identification of $ H_1 \otimes H_2 $ and $ HS(H_1^*,H_2) $, which can be proven to be isometric and isomorphic.
answer 2: For Hilbert spaces of functions, such as $ L^2(\mathbb{R}) $, you would build a tensor product as
$$ L^2(\mathbb{R}) \otimes L^2(\mathbb{R}) = L^2(\mathbb{R}^2), \qquad (\phi_1 \otimes \phi_2)(x_1,x_2) = \phi_1(x_1) \phi_2(x_2).$$
Note that the 2 arguments $ x_1, x_2 $ are different. So rather than multiplying the functions $ \phi_1, \phi_2 $ in an $x$-$y$-diagram, you go to a 3-dimensional $x_1$-$x_2$-$y$-diagram and multiply the functions
$$ f_1(x_1,x_2) = \phi_1(x_1) \; 1(x_2) \qquad \text{and} \qquad f_2(x_1,x_2) = 1(x_1) \; \phi_2(x_2) $$
(meaning $ (\phi_1 \otimes \phi_2)(x_1,x_2) = f_1(x_1,x_2) f_2(x_1,x_2) $).
More generally, $ L^2(\mathbb{R}^n) \otimes L^2(\mathbb{R}^m) = L^2(\mathbb{R}^{n+m}) $ for any $ n,m \in \mathbb{N} $, so the tensor product basically "adds the parameter dimensions of your functions". You may also use other Hilbert spaces, such as $ L^2(\Omega) $ with $ \Omega \subset \mathbb{R}^n $ being a nice set or even the Sobolev spaces $ H^s(\Omega), s \ge 0 $
answer 3: Each Hilbert-Schmidt operator $ K \in HS(L^2(\mathbb{R})^*, L^2(\mathbb{R})) $ can be identified with a kernel $ k(x_1,x_2) $ with $ k \in L^2(\mathbb{R}) \otimes L^2(\mathbb{R}) = L^2(\mathbb{R}^2) $.
That means, for $ \phi^*_1 \in L^2(\mathbb{R})^* $, the function $ K \phi^*_1 \in L^2(\mathbb{R}) $ is given by
$$ (K \phi^*_1)(x_2) = \int_\mathbb{R} k(x_1,x_2) \phi^*_1(x_1) \; dx_1$$
The identification of kernels $ k \in L^2(\mathbb{R}) \otimes L^2(\mathbb{R}) = L^2(\mathbb{R}^2) $ and Hilbert-Schmidt operators is one-to-one. So you may also construct $ L^2(\mathbb{R}) \otimes L^2(\mathbb{R}) $ using Hilbert-Schmidt operators on such function spaces.
Best Answer
First note that the Hilbert space tensor product $H_A\otimes H_B$ can be isometrically identified with the space of Hilbert-Schmidt operators from $H_A$ to $H_B$ via $$ x\otimes y\mapsto \langle \,\cdot\,,x\rangle y. $$ Every Hilbert-Schmidt operator $T\colon H_A\to H_B$ is compact and therefore has a singular value decomposition $$ T=\sum_{n} s_n\langle\,\cdot\,,e_n\rangle f_n $$ with ONBs $(e_n)$ of $H_A$, $(f_n)$ of $H_B$ and a positive sequence $(s_n)\in\ell^2$. Thus every $v\in H_A\otimes H_B$ has a representation of the form $$ v=\sum_{n} s_n e_n\otimes f_n $$ with $(e_n),(f_n),(s_n)$ as above.
The proofs can be found in virtually any introductory text on functional analysis.