Okay, let me see if I've understood what all this notation means, with the help of the Wikipedia article. Let $V$ be an $n$-dimensional real inner product space and let $F : V \to \text{Cl}(V)$ be a smooth function from $V$ to its Clifford algebra. Let $e_i$ be a basis of $V$ and let $e^i$ be its dual basis with respect to the inner product, so $e^i \cdot e_j = \delta_{ij}$. Then the $e_i$ define directional derivatives
$$\partial_i F : V \ni v \mapsto \lim_{t \to 0} \frac{F(v + t e_i) - F(v)}{\varepsilon} \in \text{Cl}(V)$$
and these assemble into a single gadget called the vector derivative
$$\partial_v : \text{Smooth}(V, \text{Cl}(V)) \ni F \mapsto \sum e^i \partial_i F \in \text{Smooth}(V, \text{Cl}(V)).$$
Since this is defined in terms of the Clifford product and since the Clifford product with a vector breaks up into an inner product (which lowers degree by $1$) and an exterior product (which increases degree by $1$), the vector derivative can be written as the sum of the corresponding operations (Doran never defines these but fortunately Wikipedia does)
$$\partial_v \cdot (-) : F \mapsto \sum e^i \cdot \partial_i F$$
$$\partial_v \wedge (-) : F \mapsto \sum e^i \wedge \partial_i F.$$
To check this let's test our understanding so far against the identities in (5.10) on p. 135. We have that
$$\partial_v (v \cdot w) = \sum e^i (e_i \cdot w) = w$$
which is fine. Next,
$$\partial_v v^2 = \sum e^i (2v \cdot e_i) = 2v$$
which is fine. Next, and this is important because it's notation we need to interpret to compute the trace,
$$\partial_v \cdot v = \sum e^i \cdot e_i = n$$
as expected. Okay, finally Doran's definition of the trace amounts to the following: if $f : V \to V$ is a linear map then
$$\boxed{ \text{tr}(f) = \partial_v \cdot f(v) = \sum e^i \cdot f(e_i) }$$
which is not new, this is just a slight variant of "sum of the diagonal entries." As far as I can tell the underline notation $\underline{f}$ he uses here is irrelevant; in Chapter 1 he defines this to refer to the extension of a linear map $f$ to blades via the exterior product but since we are taking the vector derivative here only the action of $f$ on vectors matters! At first I was trying to figure out if he was taking the derivative of $\det$ somehow but nope.
Whether this gives some geometric insight into the trace is unclear to me. Arnold's quote is about taking the derivative of $\det$ (which is the most geometric interpretation I'm aware of for the trace) and whether these two can be directly related is also unclear to me.
