Let's say I have a multivector $\mathbf{u}$ of $Cl_{3,0}$ with basis $\{1,e_1,e_2,e_3,e_1e_2,e_1e_3,e_2e_3,e_1e_2e_3\}$. A matrix representation is available in the form of the $2\times 2$ Pauli matrices.
$$ \mathbf{u}=a+xe_1+ye_2+ze_3+A_z e_1e_2+A_ye_1e_3+A_ze_1e_2+Ve_1e_2e_3 $$
How do I define a linear transformation $T$ on $\mathbf{u}$?
Option 1:
Define an $8\times 8$ matrix $T$ and the linear transformation as follows:
$$ \pmatrix{T_{01} & \dots & T_{08} \\ \vdots \\ T_{80} & \cdots & T_{88}} \pmatrix{a\\x\\y\\z\\A_z\\A_y\\A_z\\V} $$
Option 2:
Can we exploit the fact that $\mathbf{u}$ has a $2\times2$ matrix representation to define a linear transformation? For instance, the matrix representation of $\mathbf{u}$ using the Pauli matrices is:
$$ \mathbf{u}= \left( \begin{array}{cc} a+i A_z+i V+z & a+i A_x+A_y+x-i y \\ a+i A_x-A_y+x+i y & a-i A_z+i V-z \\ \end{array} \right) $$
The set of all multivectors over an $n$-dimensional vector space is itself a $2^n$-dimensional space, so the linear transformations on it are isomorphic to the $2^n\times2^n$ matrices, as you noticed.
Now consider these $2n$ "elementary" transformations, defined in terms of the geometric product:
$$f_k:A\mapsto e_kA$$
$$g_k:A\mapsto A^*e_k$$
where $A\mapsto A^*$ is defined by linearity and some other properties: $(A+B)^*=A^*+B^*,\;(AB)^*=A^*B^*,\;1^*=1,\;e_k^*=-e_k$; so if $A$ has grade $k$, then $A^*=(-1)^kA$.
The products of various subsets of $\{f_1,f_2,\cdots,f_n,g_1,g_2,\cdots,g_n\}$ are linearly independent transformations (though maybe it's not obvious), so the space generated by $f_k$ and $g_k$ is at least $2^{2n}$-dimensional. But it's a subspace of the space of all linear transformations on multivectors, which has the same dimension $2^{2n}=2^n\cdot2^n$, so they must be the same space.
Also, verify that $f_k\!^2=1$ (the identity transformation), $g_k\!^2=-1$, and otherwise $f_kf_l+f_lf_k=0$, $g_kg_l+g_lg_k=0$, $f_kg_l+g_lf_k=0$. This shows isomorphism with a geometric algebra over a pseudo-Euclidean space $\mathbb R^{n,n}$ (assuming your base field is $\mathbb R$).
See Lie Groups as Spin Groups, section V ("Endomorphisms of $\mathbb R^n$").