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.
Is it fair to say that an element $ u \otimes v = 1 u \otimes 1 v$ from the Tensor Product $ U \otimes V $ of Vector Spaces over F is a representative B of the class of bilinear maps from $ U \times V $ that send $(u,v)$ to (1,1)?
No, that doesn't sound right.
First, elements of the tensor product space are not themselves bilinear maps, so they're definitely not representatives of classes of such bilinear baps.
Second $(1,1)$ is not generally something that can be the output of a bilinear map $U\otimes V\to Z$, unless $Z$ is in particular the vector space $\mathbb F^2$.
Any bilinear map $B(v,w)$ that sends $(v,w)$ to $(1,1)$ will map $(\alpha u ,v) $ to $( \alpha,1) $ , etc.
No, that can't be. If $B$ is bilinear, that requires that $B(\alpha u,v)=\alpha\cdot B(u,v)$. So if $B$ sends $(u,v)$ to $(1,1)\in\mathbb F^2$, then it must send $(\alpha u,v)$ to $(\alpha,\alpha)$ rather than to $(\alpha,1)$.
A better understanding of the tensor elements would be that $u\otimes v$ represents a recipe for applying some bilinear map you don't know what is yet. The recipe says, "once you find out what the bilinear map is, apply it to the arguments $u$ and $v$".
In this view, far from being something that represents a bilinear map, an element of $U\otimes V$ is something that will eat a bilinear map and gives you something in that map's codomain.
The reasons tensors are different from just expressions with a variable ranging over bilinear maps is that equality between tensors knows about the bilinearity. So, for example $2u\otimes v = u\otimes 2v$ because when you apply a bilinear map to either $(2u,v)$ or $(u,2v)$ you're guaranteed you get the same result.
Best Answer
A bit more detailed look at what Qiaochu said. Unfortunately my answer won't really be expressed in terms of $r$ and $s$. I hope it still helps you in some way.
We know that $\Bbb{F}_p[X]\otimes \Bbb{F}_{p^n}\cong\Bbb{F}_{p^n}[X]$ and that $\Bbb{F}_{p^n}$ is a flat $\Bbb{F}_p$-module. Let us consider the short exact sequence $$0\to\Bbb{F}_p[X]\to\Bbb{F}_p[X]\to\Bbb{F}_p[X]/\langle r\rangle\to0,$$ where the first map is multiplication by $r$, and the last module is isomorphic to $\Bbb{F}_{p^m}$. Upon tensoring with $\Bbb{F}_{p^n}$ this gives rise to the short exact sequence $$0\to\Bbb{F}_{p^n}[X]\to\Bbb{F}_{p^n}[X]\to\Bbb{F}_{p^n}[X]/\langle r\rangle\to0.$$ Therefore a comparison of the last modules shows that $$ \Bbb{F}_{p^m}\otimes \Bbb{F}_{p^n}\cong \Bbb{F}_{p^n}[X]/\langle r\rangle. $$
The polynomial $r$ has no multiple zeros in $\overline{\Bbb{F}_p}$, so over $\Bbb{F}_{p^n}$ it factors into a product of distinct factors $$ r=\prod_{i=1}^t r_i $$ for some irreducible polynomials $r_i\in\Bbb{F}_{p^n}[X]$. Because these factors are distinct, the Chinese remainder theorem tells us that $$ \Bbb{F}_{p^n}[X]/\langle r\rangle\cong\bigoplus_i \Bbb{F}_{p^n}[X]/\langle r_i\rangle. $$
Note that everything above applies equally well to any finite extension of fields $L/K$. There is no need for the fields $L,K$ to be finite. We only needed the polynomial $r$ to be separable, so that we avoided the possibility of repeated factors.
The next step is, as Qiaochu pointed out, specific to Galois extensions. Namely, we can also deduce that the factors $r_i$ are Galois conjugates of each other. Most notably they all have the same degree. In the case of finite fields we can see this more concretely, because we know that the Galois group consists of powers of the Frobenius automorphism $F:x\mapsto x^p$. The zeros of $r$ are $$ \alpha,\alpha^p,\alpha^{p^2},\ldots,\alpha^{p^{m-1}} $$ where $\alpha$ is some (fixed) zero of $r$. For example $\alpha=X+\langle r\rangle$. The roots of one of the factors $r_i$ are then lists like $$ \alpha^{p^i},\alpha^{p^{i+n}},\alpha^{p^{i+2n}},\ldots $$ because we get such lists of conjugates by applying powers of $F^n$ to one of them.
The original list of $m$ roots consisted of a single orbit of the Galois group $G=\langle F\rangle$. This list is now partitioned into orbits of the subgroup $H=\langle F^n\rangle$. Basic facts about actions of cyclic groups tells us that the $H$-orbits all have size $m/\gcd(m,n)$, and that there are $\gcd(m,n)$ of them. Therefore we get $$ \Bbb{F}_{p^m}\otimes\Bbb{F}_{p^n}\cong\bigoplus_{i\in D}\Bbb{F}_{p^n}(\alpha^{p^i}), $$ where the set $D=\{0,1,\ldots,\gcd(m,n)-1\}$ consists of representatives of those orbits. It is easy to see that all those fields $$\Bbb{F}_{p^n}(\alpha^{p^i})\cong \Bbb{F}_{p^\ell}$$ with $\ell=\operatorname{lcm}(m,n)$.