The complex case is easier so let's start there first. We associate to a complex number $z = a + bi$ the $2 \times 2$ real matrix
$$M(z) = \left[ \begin{array}{cc} a & b \\ -b & a \end{array} \right];$$
abstractly this corresponds to considering the action of $\mathbb{C}$ on itself by left multiplication. The determinant of this matrix is the square norm $a^2 + b^2$ and we'd like a more conceptual explanation of this. The determinant is the product of the eigenvalues of $M(z)$, so what are these eigenvalues? They are exactly the complex number $z$ and its conjugate $\bar{z}$, where the corresponding eigenvectors are $\left[ \begin{array}{c} 1 \\ i \end{array} \right]$ and $\left[ \begin{array}{c} 1 \\ -i \end{array} \right]$.
In turn, one way to see why these must be the eigenvalues is to consider the characteristic polynomial of $M(z)$. This has to be a real quadratic polynomial satisfied by $M(z)$, and hence satisfied by $z$. But clearly if $z$ is not real then its minimal polynomial over $\mathbb{R}$ is
$$(t - a)^2 + b^2 = (t - z)(t - \bar{z}) = 0$$
which is exactly the real polynomial with roots $z, \bar{z}$ and so, by comparing degrees, must be the characteristic polynomial. Since the non-real $z$ are Zariski dense this must be the characteristic polynomial for all $z$. So the determinant is the squared norm $\det M(z) = z \bar{z}$ as desired.
The quaternionic case is similar. We associate to a quaternion $q = a + bi + cj + zk$ a $2 \times 2$ complex matrix $M(q)$ which I don't want to write out explicitly but which abstractly comes from considering the action of $\mathbb{H}$ on itself by left multiplication, together with the complex structure given by right multiplication by any copy of $\mathbb{C}$ inside $\mathbb{H}$, say the copy $\{ a + bi \}$ for concreteness. Again the determinant $\det M(q)$ must be the product of the eigenvalues of $M(q)$. So, what are these eigenvalues?
Again let's consider the characteristic polynomial of $M(q)$. This has to be a complex quadratic polynomial satisfied by $M(q)$, and hence satisfied by $q$ (again, we need to fix a copy of $\mathbb{C}$ inside $\mathbb{H}$, and we are taking $\{ a + bi \}$). But again if $q$ is not real then its minimal polynomial over $\mathbb{R}$ is
$$(t - a)^2 + b^2 + c^2 + d^2 = (t - q)(t - \bar{q}) = 0$$
and if $q$ is neither real nor complex then this must also be its minimal polynomial over $\mathbb{C}$, so must be the characteristic polynomial of $M(q)$. And again since the non-complex $q$ are Zariski dense this must be the characteristic polynomial for all $q$. So the determinant is the squared norm $\det M(q) = q \bar{q}$ again as desired.
Note that ultimately in the quaternionic argument we work with the minimal polynomial over $\mathbb{R}$ and end up not having to talk about $\mathbb{C}$ after all, and in both arguments we ultimately end up not really having to discuss the matrix representations at all. This whole situation is clarified enormously by knowing that there is an abstract definition of the characteristic polynomial of an element of a finite-dimensional algebra (it is "the generic minimal polynomial" in a sense which can be made precise), which generalizes the familiar case of matrix algebras and which does not depend on a choice of embedding of the algebra into a matrix algebra, and which reproduces the characteristic polynomials above. The only exposition I know of it anywhere is in Skip Garibaldi's The characteristic polynomial and determinant are not ad hoc constructions, which is well worth a read.
Best Answer
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.