Minor correction: the inclusion of the vectors is (a, b) ⇒ (0, a, b, 0), not (1, a, b, 0). Also, you have an inclusion of the scalars, which is x ⇒ (x, 0, 0, 0). And yeah, G^2 can't be projected back onto R^2 without losing a lot of information.
> what does a vector in G^2 actually represent?
If you can write a multivector x as the wedge product of k independent vectors (or equivalently, the geometric product of k orthogonal vectors), then x is called a k-blade. In this case, x has a very clear geometric interpretation:
* the magnitude of x is the volume of the parallelotope spanned by the vectors.
* the "direction" of x corresponds to the space spanned by the vectors — in concrete terms, that space is the set of vectors in R^2 that the map (v ⇒ v ∧ x) takes to zero.
* the "direction" also gives an orientation of the subspace. Orientations are defined indirectly: if you have two blades representing the same space, then they must be scalar multiples of one another. If that scaling factor is positive, then they have the same orientation, and if it is negative, then they have opposing orientations.
These three pieces of information are enough to uniquely identify x. Unfortunately, I'm not sure how to go from there to interpreting other multivectors, which are sums of blades.
> If so, then in R^3 xy, xz any yz represent rotations in the respective planes?
Yes, xy represents rotation on the xy plane, as long as you apply it as described in the article.
> Which leaves a bonus puzzle: what does (multiplying with) xyz represent?
This one should be fairly easy to prove: it represents a kind of dualization, in which the space represented by mi = mxyz is the orthogonal complement of the space represented by m.