$\DeclareMathOperator\Cl{Cl}\DeclareMathOperator\char{char}$Not sure about i), but at least ii) is true, so the isomorphism is canonical (at least over $\mathbb{R}$ or $\mathbb{C}$). To see this it is enough to show that the morphism $\pi\colon \Cl(V)\to k$ mentioned in the question is actually canonical. Indeed, if $\char(k)=0$ there is a canonical linear isomorphism $ \bigwedge^\bullet V\to \mathrm{Cl}(V)$ given by
$$
v_1\wedge v_2\wedge\dotsb \wedge v_k\mapsto \frac{1}{k!}\sum_{\sigma\in S_k}(-1)^\sigma v_{\sigma(1)}\cdot v_{\sigma(2)}\dotsm v_{\sigma(k)}.
$$  
Via this isomophism, $\Cl(V)$ inherits a natural grading (as a vector space, *not* as an algebra!), whose degree zero component is the 1-dimenional subspace generated by the unit element of $\Cl(V)$. We have therefore a distinguished canonical isomorphism $\Cl(V)^0\cong k$ and the projection on the degree zero component $\pi_0\colon \Cl(V)\to \Cl(V)^0$ is consequently a canonical $k$-linear map $\pi_0\colon \Cl(V)\to k$. It is now a simple check to see that, for any orthogonal basis $(e_1,e_2,\dotsc, e_n)$ of $V$ the linear isomorphism $ \bigwedge^\bullet V\to \Cl(V)$ maps $e_{i_1}\wedge e_{i_2}\wedge \cdots  e_{i_k}$ to $e_{i_1}\cdot e_{i_2} \cdots  e_{i_k}$, so that independently of the choice an orthonormal basis of all of the elements $g\in G^+$ with $g\neq 1$ have strictly positive degree. The a priori basis dependent map $\pi$ mentioned in the question is therefore nothing but the canonical projection on the the degree zero component $\pi_0$.

(The canonical linear isomorphism $\bigwedge^\bullet V\to \Cl(V)$ actually exists for any characteristic different from 2, but the formula is not as nice as in the characteristic zero case: one shows that the linear isomorphism induced by the basis bijection $e_{i_1}\wedge e_{i_2}\wedge \dotsb  \wedge e_{i_k} \leftrightarrow e_{i_1}\cdot e_{i_2} \dotsm e_{i_k}$ is independent of the choice of an orthogonal basis $(e_1,e_2\dotsc, e_n)$)