1) You have $5$ column vectors in $\mathbb{R}^4$ whose dimension is $4$.
So they must be linearly dependant, for every $k$.
2) Yes, $(4x^3,3x^2,2x,1)$ span $P_3$, since you can write every polynomial of dergree $3$ or less:
$$
a_3x^3+a_2x^2+a_1x+a_0=\frac{a_3}{4}4x^3+\frac{a_2}{3}3x^2+\frac{a_1}{2}2x+a_0\cdot 1.
$$
3) For your other set of polynomials, note that there are only three of them.
So their span has dimension $3$ or less.
Therefore it can not be equal to $P_3$ whose dimension is $4$.
When you say $c = ab$ is a bivector, you're taking $a \perp b$ implicitly. The geometric product will in general have both terms: $ab = a \cdot b + a \wedge b$, remember? A scalar and a bivector.
It is, admittedly, pretty hard to visualize what that is geometrically. Adding vectors to vectors or bivectors to bivectors is sensible; adding vectors and bivectors strains thinking (and I know; I tried to explain this at a talk and people repeatedly insisted it was nonsense, but how is adding real and imaginary numbers any more sensible?).
Beyond that, it is possible to think about the generalized wedge and dot products separately and to imagine a geometric interpretation. The wedge product of a vector and a bivector produces the trivector with magnitude of the parallelepiped, as you know. The dot product produces the vector in the plane that is orthogonal to the first vector.
You can think of the geometric product, then, as describing a decomposition: given a bivector $B$ and a vector $v$, $v \wedge B = v_\perp B$ extracts the part of the vector orthogonal to the surface and builds a volume from that. $v \cdot B = v_\parallel B$ extracts the part of the vector that lies in the plane and gives the orthogonal vector in the plane that could be used to reconstruct $B$ with.
You talk about sweeping vectors over each other to form the wedge product, and that's a sensible way of looking at things, but what image do we need to imagine the dot product of a vector and a bivector? I can picture taking the bivector like a flat disc of clay, putting my hands on it, and squeezing it along the direction of $v_\parallel$. That's the best I can conceive of it. That's what the dot product is, after all--a reduction operation, one that forms vectors from planes and planes from volumes.
Thinking of the dot product as "division" does seem like a sensible interpretation.
Best Answer
Both are constructed as important quotients of the tensor algebra over a vector space $V$. The polynomials are heavily linked to the symmetric algebra (via a duality), which is basically "the tensor algebra where you've injected commutativity". The exterior algebra is "the tensor algebra into which you've injected anticommutativity"; a Clifford algebra is a quantization of the exterior algebra. There are interesting morphisms between all algebras of this bunch.
The property (analogy) I think that you're noticing is that of a graded (or more generally filtered) algebra. As for their multiplication, it is what we call a discrete convolution (bilinear/distributive multiplication): that's because they're all quotients of the tensor algebra which is the "most general" way of defining a bilinear multiplication over vectors.
You may also want to look at Weyl algebras, which are quantizations of the symmetric algebra.