Is that all that the tensor product is?
Is ... is what all that the tensor product is? The previous sentence about single-qubit gates doesn't mention tensor products, so what does "that" refer to in the highlighted sentence?
If $A$ and $B$ are real algebras, then $A\otimes B$ is spanned by elements of the form $a\otimes b$, subject to the distributive property and scalar multiplication. The $\mathbb{R}$ in the subscript of $\otimes$ means only real scalars are allowed to pass across the $\otimes$ symbol, as in $(\lambda a)\otimes b=a\otimes(\lambda b)=\lambda(a\otimes b)$ for $\lambda\in\mathbb{R}$.
In the case of $\mathbb{H}\otimes_{\mathbb{R}}\mathbb{C}$, yes we are considering $\mathbb{H}$ and $\mathbb{C}$ as real algebras.
If we view $\mathbb{H}$ as a right $\mathbb{C}$-vector space, then we can multiply it by scalars from $\mathbb{H}$ on the left and by scalars from $\mathbb{C}$ on the right - these actions commute with each other because $\mathbb{H}$ is associative - which makes $\mathbb{H}$ a module over the algebra $\mathbb{H}\otimes\mathbb{C}$ as a right $\mathbb{C}$-vector space, so we have a $\mathbb{R}$- algebra homomorphism
$$ \mathbb{H}\otimes\mathbb{C}\to\mathrm{End}_{\mathbb{C}}(\mathbb{H}) $$
Note $\mathbb{H}\cong\mathbb{C}^2$ as a $\mathbb{C}$-vector space so $\mathrm{End}_{\mathbb{C}}(\mathbb{H})\cong M_2(\mathbb{C})$. We can check the above is an isomorphism; pick basis elements $\{1,\mathbf{i},\mathbf{j},\mathbf{k}\}$ and $\{1,i\}$ of $\mathbb{H}$ and $\mathbb{C}$ to form basis elements $a\otimes b$ of $\mathbb{H}\otimes\mathbb{C}$, then check the corresponding matrices in $M_2(\mathbb{C})$ are linearly independent. The homomorphism (turning tensor into matrices) is a bit tricky because we're combining left/right actions...
Here's how to turn $\mathbf{j}\otimes i$ into a $2\times 2$ complex matrix. First, for $\mathbb{H}\cong\mathbb{C}^2$ as right $\mathbb{C}$-vector spaces, we'll use $\{1,\mathbf{j}\}$ as a basis corresponding to $(1,0)$ and $(0,1)$. Then we can define $(a\otimes b)x:=axb$ (you could also define $ax\overline{b}$ instead, the conjugation being useful for ensuring we get a left module in general, but it won't matter here because $\mathbb{C}$ is commutative). So we compute
$$ (\mathbf{j}\otimes i)(1) ~=~ \mathbf{j}i ~=~ 1(0+0i)+\mathbf{j}(0+1i) $$
$$ (\mathbf{j}\otimes i)(\mathbf{j}) ~=~ \mathbf{jj}i ~=~ 1(0-1i)+\mathbf{j}(0+0i)$$
So the matrix is
$$ \mathbf{j}\otimes i \quad\longleftrightarrow\quad \begin{bmatrix} 0 & -i \\ i & 0 \end{bmatrix} $$
You mean use a different convention for how $\mathbb{H}$ is a $\mathbb{H}\otimes\mathbb{C}$-module in order to be more consistent with the notes you're using.
It is unclear what is unclear to you.
I am saying that $\Bbb{H}$ is just a 2-dimensional (left) vector space so we can replace $End_\Bbb{C}(\Bbb{H})$ by $End_\Bbb{C}(\Bbb{C}^2)$.
It means that I am interpreting $End_\Bbb{C}(\Bbb{H})$ as $$End_{left\ \Bbb{C}\ vector\ space}(\Bbb{H})$$
Then $1\otimes a+i\otimes b+j\otimes c+ij\otimes d\mapsto a\pmatrix{1&0\\0&1}+b\pmatrix{i&0\\0&-i}+c\pmatrix{0&1\\-1&0}+d\pmatrix{0&i\\i&0}$ gives the isomorphism $\Bbb{H\otimes_R C }\to End_\Bbb{C}(\Bbb{C}^2)$.
The $i$ of $i\otimes b$, the $i$ hidden in $a=A+iB$, and the $i$ in $\pmatrix{i&0\\0&-i}$ are not the same, that's part of the game to understand what we identity the latter two.
The distinction between left and right vector space is because $\Bbb{C}$ is a subring of $\Bbb{H}$ but which doesn't commute with $\Bbb{H}$.
Best Answer
In this context, an "algebra" is short for "$F$-algebra" for some field $F$. When we say $A$ is an $F$-algebra, that means the multiplication on $A$ is $F$-bilinear. In particular, this requires $F$ to be commutative (not a skew field like $\mathbb{H}$) and central (so $\mathbb{H}$ is not a $\mathbb{C}$-algebra since $Z(\mathbb{H})=\mathbb{R}$). Things go wrong if you try to loosen this definition.
Note, by the way, you can change what field you consider an algebra over. For example, $A=M_2(\mathbb{C})$ can be considered as an $\mathbb{R}$-algebra or a $\mathbb{C}$-algebra. Facts like (non)commutativity and (non)associativity are properties of $A$'s multiplication operation and do not depend on which $F$ is chosen.
Suppose $M$ and $N$ are right and left $A$-modules, respectively. Then $M\otimes_AN$ is a well defined $F$-vector space. If actually $M$ is a $(X,A)$-bimodule (meaning left $X$-module, right $A$-modules, and these module actions commute) and $N$ is an $(A,Y)$-module, then $M\otimes_AN$ will be a $(X,Y)$-bimodule. (This is only a gain in generality, not a loss, because we can take $X$ and/or $Y$ to be $F$.)
Note $\mathbb{O}$ is only an $\mathbb{R}$-algebra (unital, nonassociative), not a $\mathbb{C}$-algebra or $\mathbb{H}$-algebra. It is a left $\mathbb{C}$-vector space and a right $\mathbb{C}$-vector space (i.e. module), indeed a $(\mathbb{C},\mathbb{C})$-bimodule because it is an alternative algebra, however it is not a left or right $\mathbb{H}$-vector space (let alone a bimodule).
That said, you can still define "$\mathbb{O}\otimes_\mathbb{H}\mathbb{O}$" as a vector space to be the quotient of $\mathbb{O}\otimes_\mathbb{R}\mathbb{O}$ by the vector subspace spanned by the differences $(xh\otimes y-x\otimes hy)$ for $x,y\in\mathbb{O}$, $h\in\mathbb{H}$. Using $\ell$ for any nonzero imaginary octonion perpendicular to $\mathbb{H}$, we can say $\mathbb{O}=\mathbb{H}\oplus\mathbb{H}\ell$ is a $\mathbb{Z}_2$-graded vector space, with $\ell\mathbb{H}=\mathbb{H}\ell$, and then $\mathbb{O}\otimes_{\mathbb{R}}\mathbb{O}=(\ell\mathbb{H}\oplus\mathbb{H})\otimes_{\mathbb{R}}(\mathbb{H}\oplus\mathbb{H}\ell)$ becomes $\mathbb{Z}_2\times\mathbb{Z}_2$-graded:
$$ (\ell\mathbb{H}\otimes\mathbb{H}) \,\oplus\, (\ell\mathbb{H}\otimes\mathbb{H}\ell) \,\oplus\, (\mathbb{H}\otimes\mathbb{H}) \,\oplus\, (\mathbb{H}\otimes\mathbb{H}\ell). $$
(All unadorned $\otimes$ are over $\mathbb{R}$.) Notice the subspace we're quotienting by is also $\mathbb{Z}_2\times\mathbb{Z}_2$-graded, so we can quotient in each of the four "components" separately. Notice $\ell x\otimes y=(\ell x)y\otimes1=\ell(yx)\otimes1$ but also $\ell x\otimes y=\ell\otimes xy=\ell(xy)\otimes 1$, therefore the difference $\ell z\otimes 1=0$ vanishes, where $z=xy-yx=$ $2\,\mathrm{Im}(x)\times\mathrm{Im}(y)$ can be any pure imaginary octonion. The same cancellation occurs with $\ell$ on the right of $\otimes$, but with $\ell$ on both or neither side no cancellation occurs (but quaternion scalars can be slid to once side).
Thus the four $\mathbb{Z}_2\times\mathbb{Z}_2$-components become, in the quotient,
$$ \mathbb{O}\otimes_{\mathbb{H}}\mathbb{O} = \mathbb{R}(\ell\otimes1) \,\oplus\, (\ell\otimes\mathbb{H}\ell) \,\oplus\, (1\otimes\mathbb{H}) \,\oplus\, \mathbb{R}(1\otimes\ell). $$
So it would appear this "$\mathbb{O}\otimes_{\mathbb{H}}\mathbb{O}$" is a $10$-dimensional $\mathbb{R}$-vector space. There is a natural map $\mathbb{O}\otimes_{\mathbb{R}}\mathbb{O}\to\mathbb{O}\otimes_{\mathbb{H}}\mathbb{O}$ given by $a\otimes b\mapsto a\otimes b$, but the algebra's multiplication map $\mathbb{O}\otimes_{\mathbb{R}}\mathbb{O}\to\mathbb{O}$ does not factor through it like it would in the associative world.