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.
The Hilbert tensor product is in general not equal to the projective tensor product:
If $H$ is a Hilbert space and $H^*$ its dual space,
then
- $H \hat \otimes_\pi H^*$ (the projective tensor product) is (isometrically isomorphic to)
the trace class (nuclear) operators with the trace norm
- $H \hat \otimes_\epsilon H^*$ (the injective tensor product) is (isometrically isomorphic to)
the compact operators with the operator norm
- $H \hat \otimes_h H^*$ (the Hilbert tensor product, which is a Hilbert space again) is (isometrically isomorphic to) the Hilbert-Schmidt operators with the Hilbert-Schmidt norm
Since the spaces of Hilbert-Schmidt, compact and trace class operators are, in infinite dimensions, never the same it follows that the tensor norms can not be the same either.
To show that they are never the same in infinite dimensions:
Let $(e_n)_{n \in \mathbb{N}}$ be an orthonormal system in $H$.
Let $(x_n)_{n \in \mathbb{N}}$ be any sequence of complex numbers that converges to 0, but whose absolute value is not square summable. For example $x_n =1/\sqrt{n}$.
Define a linear operator $T: H\to H$ by $Ty = \sum_{n \in \mathbb{N}} x_n \langle e_n, y \rangle e_n$. Then $T$ is compact, but not Hilbert-Schmidt.
Let $(s_n)_{n \in \mathbb{N}}$ be any sequence of complex numbers whose absolute value is square summable, but not summable.
For example $s_n := 1/n$.
Define a linear operator $S : H \to H$ by $Sy = \sum_{n \in \mathbb{N}} s_n \langle e_n, y \rangle e_n$. Then $S$ is Hilbert-Schmidt, but not trace class.
Best Answer
Once you have an isometry between two pre-Hilbert spaces $A$ and $B$, you know that the resulting Hilbert spaces $\overline{A}$ and $\overline{B}$ are isometric as well, since your isometry will canonically extend to the completions.
However, the Hilbert space $(H_1\otimes H_2)\otimes H_3$ is by definition the completion of the pre-Hilbert space $(H_1\otimes H_1)\odot H_3$, not of $(H_1\odot H_2)\odot H_3$.
Hence, what you need to check is that going from $(H_1\odot H_2)\odot H_3$ to $(H_1\otimes H_2)\odot H_3$ by completing the first factor and then to $(H_1\otimes H_2)\otimes H_3$ by completing the result yields a Hilbert space that is canonically isometric to what you get from directly completing the pre-Hilbert space $(H_1\odot H_3)\odot H_3$.
So, in your attempt, the extension of $\psi$ to completions should be $$ \overline \psi \colon \underbrace{H_1\otimes H_2\otimes H_3}_{= \overline{H_1\odot H_2\odot H_3}} \stackrel{\cong}\longrightarrow \overline{(H_1\odot H_2)\odot H_3} $$ and you are missing an isometry $$ \overline{(H_1\odot H_2)\odot H_3} \cong \underbrace{(H_1\otimes H_2)\otimes H_3}_{=\overline{\left(\overline{H_1\odot H_2}\right)\odot H_3}}. $$
See Proposition 2.6.5 in R.V. Kadison, J. R. Ringrose: Fundamentals of the Theory of Operator Algebras (1983) for a general proof of $$H_1\otimes \dots \otimes H_{n+m} \cong (H_1\otimes \dots\otimes H_n)\otimes(H_{n+1}\otimes\dots\otimes H_{n+m})$$ using a universal property of tensor products of Hilbert spaces.