It would really helpful if the article had some worked examples of this arithmetic, with actual numbers, the kind with digits and decimal points in. Then, things like the structure of a geometric product would be absolutely clear.
Let me have a go. I'll use an asciified version of the symbols, with * to mean multiplication of two scalars,, to mean raising one scalar to the power of another, and _ to mean taking a component of a vector.
ab = a.b + a^b
a.b = a_x * b_x + a_y * b_y + a_z * b_z
a^b = (a_x * b_y - b_x * a_y) (x^y)
+ (a_x * b_z - b_x * a_z) (x^z)
+ (a_y * b_z - b_y * a_z) (y^z)
ab = (a_x * b_x + a_y * b_y + a_z * b_z)
+ (a_x * b_y - b_x * a_y) (x^y)
+ (a_x * b_z - b_x * a_z) (x^z)
+ (a_y * b_z - b_y * a_z) (y^z)
So if a = (1, 2, 3) and b = (4, 5, 6):
ab = (1 * 4 + 2 * 5 + 3 * 6)
+ (1 * 5 - 4 * 2) (x^y)
+ (1 * 6 - 4 * 3) (x^z)
+ (2 * 6 - 5 * 3) (y^z)
= 32 + -3 (x^y) + -6 (x^z) + -3 (y^z)
The dot product makes a scalar, the wedge product makes a bivector, and the geometric product makes a scalar plus a bivector
You will note that the scalar part is much bigger than the coefficients of the bivector part. That's because the input vectors are actually quite similar - pointing z-by-y-z, with a little bit of x. Hence, their projection onto each other is large, whereas the parallelogram they form is quite small (long and thin). The dot product measures the former, the wedge product the latter.
Have i got that right?
EDIT And to clarify this:
> For any basis vector, such as the x axis, the result [of taking the geometric product with itself] is 1
That '1' isn't the scalar number 1, it's the scalar-plus-bivector 1 + 0 (x^y) + 0 (x^z) + 0 (y^z).