Let us say that I have a set $\mathbb{V}$ of multivectors of $Cl_4(\mathbb{C})$. They are expressed as follows:
$$ \mathbf{v}_1=R_1+ t_1 \mathbf{e}_0 + x_1 \mathbf{e}_1 +y_1 \mathbf{e}_2 +z_1 \mathbf{e}_3 \\ \vdots\\ \mathbf{v}_n=R_n+ t_n \mathbf{e}_0 + x_n \mathbf{e}_1 +y_n \mathbf{e}_2 +z_n \mathbf{e}_3 $$
I 'visualize' each element of $\mathbb{V}$ as assigning a scalar value $R_i$ to each point in space-time. Furthermore, 'metric information' is encoded in the basis $\{\mathbf{e}_0,\mathbf{e}_1,\mathbf{e}_2,\mathbf{e}_3\}$. Consequently, since it appears to have all the required ingredients, $\mathbb{V}$ reminds me of a field in physics.
Fields in physics are usually defined as $R:(x,y,z,t)\to \mathbb{R}$, and an integral over such a field is usually defined as $I=\int R\sqrt{g}d^4x$.
Is $\mathbb{V}$ analogous to a field?
I am trying to find the logical equivalent to $I=\int R\sqrt{g}d^4x$ but instead using $\mathbb{V}$ as a starting point.
Specifically, how do I sum over all elements of $\mathbb{V}$ to obtain a field integral?
I think that your construction is over-specified. You want 4 parameters to parameterize the space time volume. For example, suppose that you have a vector function of time and space coordinates that spans the integration volume of interest $$\mathbf{u} = t \mathbf{e}_0 + x \mathbf{e}_1 + y \mathbf{e}_2 + z \mathbf{e}_3,$$ then the natural way to express the differential volume element is just $$ \sqrt{g} d^4 x = \left( { \frac{\partial {\mathbf{u}}}{\partial {t}} dt } \right) \wedge \left( { \frac{\partial {\mathbf{u}}}{\partial {x}} dx } \right) \wedge \left( { \frac{\partial {\mathbf{u}}}{\partial {y}} dy } \right) \wedge \left( { \frac{\partial {\mathbf{u}}}{\partial {z}} dz } \right) = \mathbf{e}_0 \mathbf{e}_1 \mathbf{e}_2 \mathbf{e}_3 dt dx dy dz.$$ I've included your $ \sqrt{g} $ term on the left hand side since the wedge product of the differentials implicitly builds in the Jacobian structure that I assume this represents.