$
\newcommand\Cl{\mathrm{Cl}
\newcommand\tr{\mathop{\mathrm{tr}}}}
\newcommand\Ext{{\textstyle\bigwedge}}
\newcommand\form[1]{\langle#1\rangle}
\newcommand\Hom{\mathop{\mathrm{Hom}}}
\newcommand\rev[1]{#1^\perp}
\newcommand\doub\mathfrak
$

Suppose $k$ has characteristic $\not= 2$ and that $V$ is finite dimensional (which I think you assumed implicitly). Let $V^*$ be the dual of $V$. We will also need to assume that $Q$ is nondegerate, i.e. its bilinear form gives an isomorphism $V \cong V^*$.

Then $\dim\Cl(V, Q) = 2^n$; if we define the trace of an element of $\Cl(V, Q)$ as the trace of its left multiplication
$$
  \tr X = \tr(Y \mapsto XY)
$$
then your $\pi : \Cl(V, Q) \to k$ is exactly
$$
  \pi(X) = \frac1{2^n}\tr X.
$$
This can be confirmed by using an orthonormal basis. Alternatively, we could construct the canonical isomorphism $\Cl(V, Q) \cong \Ext V$ where then
$$
  \pi(X) = \form{X}_0
$$
where the RHS is the projection onto the scalar (i.e. grade 0) part imported from $\Ext V$ into $\Cl(V,Q)$. Briefly, we can do this by defining $\wedge : V \times \Cl(V,Q) \to \Cl(V,Q)$ by $v\wedge X = \frac12(vX + \hat Xv)$ where $\hat X$ is the main involution applied to $X$ (i.e. the involution that negates all vectors); considering the map $V \to \Hom_k\Cl(V,Q)$ given by $v \mapsto v\wedge\cdot$, the universal property of $\Ext V$ extends this to an algebra homomorphism $\varphi : \Ext V \to \Hom_k\Cl(V, Q)$, and the map $X \mapsto \varphi(X)(1)$ can be shown to be the desired isomorphism. This is a particular instance of Chevalley's maps $\Cl(V, Q) \cong \Cl(V, Q+Q')$; see [these notes](https://www.cip.ifi.lmu.de/~grinberg/algebra/chevalleys.pdf) by Darij Grinberg for a general approach descending from the tensor algebra.

The bilinear form associated to $Q$
$$
  B(v, w) = \frac12(Q(v + w) - Q(v) - Q(w))
$$
induces an isomorphism $\flat : V \to V^*$; using this to apply $Q$ to $V^*$, the universal properties of the Clifford algebras $\Cl(V, Q), \Cl(V^*, Q)$ extend $\flat$ to an algebra isomorphism $\Cl(V, Q) \to \Cl(V^*, Q)$. Using the natural bilinear pairing on $\Ext V^*\times\Ext V \to k$
$$
  (v^*_l\wedge v^*_{l-1}\wedge\cdots\wedge v^*_1,\; v_1\wedge v_2\wedge\cdots\wedge v_m) \mapsto \delta_{lm}\det\bigl(v^*_i(v_j)\bigr)_{i,j=1}^m
$$
then induces a linear isomorphism $\Ext V^* \to (\Ext V)^*$, and composing with $\flat$ finally gives a linear isomorphism $\flat' : \Cl(V, Q) \to \Cl(V, Q)^*$; the bilinear form associated to this turns out to be exactly
$$
  (X, Y) \mapsto \form{XY}_0.
$$
We can actually get the pairing $\Ext V^*\times\Ext V \to k$ by considering the natural Clifford algebra on $V^*\oplus V$ and using the trace definition of the scalar part, but I will not digress further.

Now we define the isomorphism $M^\vee \to M^*$ by
$$
  \psi \mapsto \Phi_\psi,\quad \Phi_\psi(m) = \form{\psi(m)}_0.
$$
This is a homomorphism since $\form{XY}_0 = \form{YX}_0$ and so
$$
  \Phi_{\psi\cdot g}(m) = \form{\psi(m)g}_0 = \form{g\psi(m)}_0 = \form{\psi(gm)}_0 = (\Phi_\psi\cdot g)(m).
$$
The inverse is given by
$$
  \phi \mapsto \Psi_\phi,\quad \Psi_\phi(m) = (g \mapsto \phi(gm))^{\sharp'}
$$
where $\sharp' = (\flat')^{-1}$. Observe:
$$
  \Psi_{\Phi_\psi}(m)
    = (g \mapsto \Phi_\psi(gm))^{\sharp'}
    = (g \mapsto \form{\psi(gm)}_0)^{\sharp'}
    = (g \mapsto \form{g\psi(m)}_0)^{\sharp'}
    = \psi(m).
$$

Your formula in terms of $G^+$ comes from expressing $\sharp'$ using that basis; the reciprocal basis of $G^+$ is $\{g^{-1} \;:\; g \in G^+\}$.

---

I have no idea how to handle characteristic 2. The main issue is that there is no canonical isomorphism with $\Ext V$ in this case, and we can't even use the trace since $\tr X = 0$ for all $X$.