I have been reading Lawson & Michelsohn's Spin Geometry, and in the beginning of Chapter $1$ $\S 4$ on classification of the Clifford algebra $\text{Cl}_{r,s}$, they say that
With little difficulty the reader can check the first few cases $$\text{Cl}_{1,0}=\mathbb C\quad\text{Cl}_{0,1}= \mathbb R \oplus\mathbb R\\\text{Cl}_{2,0}=\mathbb H\quad\text{Cl}_{0,2}=\mathbb R(2)\\\text{Cl}_{1,1}=\mathbb R(2)$$ where $\mathbb R(2)$ denotes the algebra of $2\times 2$ real matrices.
I think I understand the first three cases: for $\text{Cl}_{1,0}$, we add a basis $x_1$ such that $x_1^2=-1$ to $\text{Cl}_{0,0}=\mathbb R$, so it is $\mathbb C$ by identifying $x_1$ with $i\in\mathbb C$; for $\text{Cl}_{0,1}$, we add a basis $x_1$ such that $x_1^2=1$ to $\text{Cl}_{0,0}=\mathbb R$, so it is $\mathbb R\oplus \mathbb R$; for $\text{Cl}_{2,0}$, we add two bases $x_3,x_4$ such that $x_3^2+x_4^2=-1$ to $\text{Cl}_{1,0}=\mathbb C$, so it is $\mathbb H$ by identifying $x_3,x_4$ with $j,k\in\mathbb H$.
Questions: Is my understanding of the first three cases correct? How do I show the last two cases?
$ \newcommand\Cl{\mathrm{Cl}} \newcommand\R{\mathbb R} \newcommand\End{\mathrm{End}} $(A note for others: OP is using the convention that $\Cl_{r,s}$ has $r$ generators squaring to $-1$ and $s$ generators squaring to $1$, rather than the convention that seems to be more common nowadays where $r$ and $s$ are swapped.)
For $\Cl_{2,0}$ it should be $x_3^2 = x_4^2 = -1$. I assume this was a typo, in which case your understanding of the first three cases is correct so long as you recognize the importance of the fact that the algebras are associative and that the generators $x_1, x_2, x_3, x_4$ anticommute.
For the last two cases I can't know what the author intended, but a generally applicable idea is the following.
We identify vectors $V := \R^{r+s}$ as a subspace of $\Cl_{r,s}$ in such a way that $x_1,\dotsc,x_r,y_1,\dotsc,y_s$ is the standard basis of $V$. The algebra $\Cl_{r,s}$ is generated by $V$ and inherits a $\mathbb Z/2\mathbb Z$-grading where homogeneous elements are sums of products of an even number of vectors or sums of products of an odd number of vectors; call these even and odd elements. The even elements form a subalgebra $\Cl_{r,s}^+$, and we prove that $$ \Cl_{r-1,s} \cong \Cl_{r,s}^+ \cong \Cl_{s-1,r} $$ (so long as $s-1 \geq 0$ or $r-1 \geq 0$). Consider the even elements $$ x_2x_1^{-1},\quad x_3x_1^{-1},\quad \dotsc,\quad x_rx_1^{-1},\quad y_1x_1^{-1},\quad y_2x_1^{-1},\quad \dotsc,\quad y_sx_1^{-1}. $$ They are linearly independent since $X \mapsto Xx_1^{-1}$ is an injective linear map; they anticommute $$ zx_1^{-1}z'x_1^{-1} = -zz'x_1^{-2} = z'zx_1^{-2} = -z'x_1^{-1}zx_1^{-1} $$ (where $z, z' = x_2,\dotsc,x_r,y_1,\dotsc,y_s$ and $z \ne z'$); they satisfy $$ (x_ix_1^{-1})^2 = x_ix_1^{-1}x_ix_1^{-1} = -x_i^2x_1^{-2} = -1, $$ $$ (y_jx_1^{-1})^2 = y_jx_1^{-1}y_jx_1^{-1} = -y_j^2x_1^{-2} = 1 $$ (where $i \ne 1$); and their products give the standard basis for $\Cl_{r,s}^+$. This shows that $\Cl_{r,s}^+ \cong \Cl_{r-1,s}$. Replacing $x_1$ with $y_1$ in the above will similarly show that $\Cl_{r,s}^+ \cong \Cl_{s-1,r}$. Note that $y_1^{-1} = y_1$. The use of $x_1^{-1}$ above is a convention and is not essential.
For $y_1$ we get a representation $\rho : \Cl_{r,s} \to \End_\R(\Cl_{r,s}^+)$ defined on vectors $v$ by $$ \rho(v)(Y) = vYy_1,\quad Y \in \Cl_{r,s}^+ $$ and on all of $\Cl_{r,s}$ by extending to a homomorphism using its universal property. We prefer this representation when $s > 0$. When $s = 0$ things are trickier but this is not relevant to $\Cl_{0,2}$ and $\Cl_{1,1}$. This representation is injective since if $XYy_1 = 0$ for some $X \in \Cl_{r,s}$ and all $Y \in \Cl_{r,s}^+$ we simply choose invertible $Y$ and conclude $X = 0$.
Now we compare dimensions in the special case $r+s = 2$: $$ \dim\End_\R(\Cl_{r,s}^+) = (2^{r+s-1})^2 = 4 = 2^{r+s} = \dim\Cl_{r,s}. $$ Thus $\rho$ is an algebra isomorphism in this case. Finally, the existence of $\rho$ allows us to conclude $$ \Cl_{0,2}^+ \cong \Cl_{1,0} \cong \mathbb C \implies \Cl_{0,2} \cong \End_\R(\mathbb C) \cong \R(2), $$ $$ \Cl_{1,1}^+ \cong \Cl_{0,1} \cong \R\oplus\R \implies \Cl_{1,1} \cong \End_\R(\R\oplus\R) \cong \R(2). $$