Visualizing the geometric product? The exterior product between blades has a relatively clear geometric interpretation: it gives the result of "extending" one factor along the other, with the direction pointing along the first factor and then the second. So the exterior product of two vectors can be imagined as the directed parallelogram between two arrows.
Is there a similar way to visualize the geometric product? Or must it just be considered algebraically?
 A: The most intuitive interpretation of a Geometric Product I have found is from Hestenes who notes that it can be visualized as a directed arc just as a vector can be viewed as a directed line.

For more depth, see page 11 of the following:
http://geocalc.clas.asu.edu/pdf/OerstedMedalLecture.pdf
A: This is based on my answer given here: https://math.stackexchange.com/a/4304117/322997
Vectors can be thought of as representing (hyper)planes. For example, in 3D space a vector can be used to represent a plane through the origin. Now, a plane $u$ can be reflected in a plane $v$ using
$$ v[u] = -vuv^{-1}, $$
where the minus sign is needed such that when you reflect $v$ in itself, the front and back of the mirror flip ($v[v] = -v$). If we now also perform a reflection in a second plane $w$, we get the rotation
$$ w[v[u]] = (wv)u v^{-1} w^{-1} = (wv) u (wv)^{-1} $$
The composition of two reflections $wv$ is called a bireflection, and could in fact be either a rotation, translation, or a boost. The picture below shows how two intersecting reflections form a rotation, while parallel reflections form a translation.

So the product of two vectors is a bireflection.
The "apples and oranges" of scalar plus bivector only appear because in order to actually compute it, we would have to somehow choose a basis. Staying with the 3D example, we could choose an orthogonal basis $e_1, e_2, e_3$ such that $e_i e_j = \delta_{ij} + e_{ij}$ and represent any plane as $x = \sum_i x^i e_i$.
Now, when we compute the bireflection $wv$ we will get a scalar and bivector part:
$$ w v = \sum_{ij}(w^i e_i) (v^j e_j) = \sum_i w^i v^i + \sum_{i \neq j} w^iv^j e_{ij}. $$
So just remember the truth: there are no apples and oranges. This works in any number of dimensions: the vectors of a geometric algebra $\mathbb{R}_{p,q,r}$ form the reflection group $Pin(p,q,r)$. For more detail on this approach I would refer to this video, or to the Graded Symmetry Groups paper. Full disclaimer: I'm one of the authors.
