$
\newcommand\Cl{\mathrm{Cl}}
\newcommand\Ext{{\textstyle\bigwedge}}
\newcommand\rev\widetilde
\newcommand\rintr{\mathbin{{\llcorner}}}
\newcommand\R{\mathbb R}
\newcommand\form[1]{\langle#1\rangle}
\newcommand\dd{\mathrm{d}}
\newcommand\lcontr{\mathbin{{\rfloor}}}
\newcommand\rcontr{\mathbin{{\lfloor}}}
\newcommand\doub{\mathfrak}
\newcommand\doubnabla{\boldsymbol\nabla}
\newcommand\adj{\overline}
$
Let $Q$ be a quadratic form over an $n$-dimensional vector space $V$ (which for this discussion we will assume is real though this isn't strictly necessary.) Quadratic forms $Q$ and symmetric bilinear forms $B$ are in one-to-one correspondence via
$$
B(v, w) = \frac12(Q(v + w) - Q(v) - Q(w)),\quad Q(v) = B(v, v).
$$
In this context, metric is another word for symmetric bilinear form.
The Clifford algebra $\Cl(V, Q)$ is essentially the associative algebra generated by $V$ with the relations $v^2 = Q(v)$ for all $v \in V$. This can be formalized by defining $\Cl(V, Q)$ as the algebra with a certain universal property, or by quotienting the tensor algebra by the previously mentioned relations. It is not defined in terms of a wedge product; I would consider such a definition a bad definition.
But there are two important relationships between Clifford algebras and the exterior algebra $\Ext V$. First, $\Cl(V, 0)$ is exactly $\Ext V$, where $0$ is the trivial quadratic form. Second, there is a canonical linear isomorphism (not an algebra isomorphism) $\Cl(V, Q) \cong \Ext V$ for any $Q$. This means that we can view $\Cl(V, Q)$ as $\Ext V$ endowed with a second product, the Clifford product, and one way of constructing $\Cl(V, Q)$ is by defining such a product on $\Ext V$ (but I would never take this as a definition).
In light of this, we can view various operations on $\Ext V$ as operations on $\Cl(V, Q)$. This is very natural with e.g. the Hodge star $\star$. Denote by $\Cl^k(V, Q)$ the linear subspace of $\Cl(V, Q)$ corresponding to $\Ext^k V$. There is an antiautomorphism $A \mapsto \rev A$ of $\Cl(V, Q)$ which reverses all products and is the identity on $V$. We get the Hodge star when $Q$ is non-degenerate and when we choose $I \in \Cl^n(V, Q)$ such that $I^2 = \pm1$ (such elements always square to scalars); this is equivalent to choosing an orientation on $V$. Then the Hodge star is exactly
$$
\star A = \rev AI
$$
for any $A \in \Cl(V, Q) \cong \Ext V$.
A differential form assigns an element of $(\Ext V)^*$ to each tangent space of a manifold (where we assume the tangent spaces are isomorphic to $V$). There is a canonical isomorphism $(\Ext V)^* \cong \Ext V^*$ induced by a natural bilinear pairing $\form{{-}, {-}} : \Ext V^* \times \Ext V \to \R$, so we usually identify a differential form with an element of $\Ext V^*$. Then the action of a differential form $A^* \in \Ext V^*$ on $B \in \Ext V$ is given by $\form{A^*, B}$, or equivalently the action on vectors $v_1,\dotsc, v_k \in V$ is given by
$$
\form{A^*, v_1\wedge\cdots\wedge v_k}.
$$
But in the presence of a non-degenerate metric there is an isomorphism $V \cong V^*$, and this extends naturally to algebra isomorphisms $\Ext V \cong \Ext V^*$ and $\Cl(V, Q) \cong \Cl(V^*, Q^*)$ (where $Q^*$ is $Q$ applied to $V^*$ via the isomorphism). This means we can identify differential forms with elements of $\Ext V$. The pairing $\form{{-}, {-}}$ becomes a bilinear form on $\Ext V$, and is (almost) exactly the scalar part of the Clifford product:
$$
\form{A, B} = \form{\rev AB}_0,\quad A, B \in \Cl(V, Q) \cong \Ext V.
$$
This scalar part is defined either by taking advantage of the grading of $\Ext V$, or as intrinsic to the Clifford algebra via the normalized trace:
$$
\form{A}_0 = \frac1{2^n}\mathrm{Tr}(B \mapsto AB),\quad B \in \Cl(V, Q).
$$
The reversal $\rev A$ in the above and in the Hodge star is an artifact of how $\form{{-},{-}}$ is defined, and can be done away with if desired. For simplicity, we will use this convention and define
$$
\form{A, B} = \form{AB}_0,\quad \star A = AI.
$$
In this way, we can represent differential forms as elements of $\Cl(V, Q)$ and use all the tools that comes with that.
As a particular example, we get the exterior derivative as
$$
[\dd\form{A_x, {-}}](B) = \form{(\nabla\wedge A_x)B}_0
$$
and so the codifferential as
$$
[{\star^{-1}}\dd{\star}\form{A_x, {-}}](B)
= \form{(\nabla\wedge(A_xI))I^{-1}B}_0
= \form{(\nabla\lcontr A_x)B}_0
$$
where $\lcontr$ is the left contraction
$$
A\lcontr B = \form{AB}_{k-j},\quad A \in \Cl^j(V, Q),\quad B \in \Cl^k(V, Q),
$$
i.e. the projection onto the exterior algebra grade ${k-j}$ component, which we define to be zero if $j > k$. This is the manifestation of the interior product in the presence of a metric, as is the closely related right contraction $\rcontr$. To summarize, the exterior derivative and codifferentials are expressed in $\Cl(V, Q)$ as
$$
\nabla\wedge A_x,\quad \nabla\lcontr A_x.
$$
But for any $v \in V$ and $A \in \Cl(V, Q)$
$$
vA = v\wedge A + v\lcontr A
$$
so we see that the sum of the exterior derivative and codifferential is the geometric derivative
$$
\nabla A_x = \nabla\wedge A + \nabla\lcontr A
$$
where on the left we are using the Clifford product.
This is extremely useful; for example, Maxwell's equations
become exactly one equation $\nabla F = J$, and (at least in flat space) $\nabla$ has a Green's function
which allows us to directly solve for $F$ in terms of $J$.
This is the beginning of geometric calculus;
I would recommend chapter 6 of Doran and Lasenby's Geometric Algebra for Physicists (2003) for more.
(Really, the whole book is worth a read.)
In brief, this approach to differential geometry (of course with a focus on Riemannian geometry)
involves embedding our manifold in flat space,
where then one of the major objects of study becomes the bivector-valued shape tensor $S_x$
which encodes much of the extrinsic and intrinsic geoemtry of the manifold.
For instance, the covariant derivative (i.e. Levi-Civita connection) can be expressed as
$$
DA_x = \partial A_x + \partial_a(S_x(a)\times A_x)
$$
where $X\times Y = \tfrac12(XY - YX)$, the tangential derivative $\dot\partial = P_x(\dot\nabla)$
is the "projection" of $\nabla$ onto the tangent space of the manifold,
and $\partial_a$ is the tangential derivative with respect to $a$
(actually representing a kind of tensor contraction since $S_x$ is linear).
The introduction of a metric obviously takes us away from pure differential forms.
We can actually try to do away with the metric:
the space $\doub W = V^*\oplus V$ has a natural bilinear form
$$
\form{v^* + v, w^* + w} = v^*(w) + w^*(v)
$$
with quadratic form
$$
\form{v^* + v, v^* + v} = 2v^*(v).
$$
We then consider the Clifford algebra $\Cl(\doub W)$.
Since $\form{v^*, v^*} = \form{v, v} = 0$,
the subalgebras generated by $V^*$ and $V$ are exactly $\Ext V^*$ and $\Ext V$.
Not only that, but if $M$ is a manifold such that $T_xM \cong V$,
then the tangent spaces of the cotangent bundle
are naturally isomorphic to $\doub W$.
This makes $\Cl(\doub W)$ a potentially very natural setting in which to study differential forms.
However, I have yet to study this construct to my satisfaction, so I will stop here.
