Let $\|\cdot\|_{d_g} : L^\infty(M) \to [0,+\infty]$ be the *Lipschitz seminorm* induced by the geodesic metric $d_g$ on $M$, defined by
$$
\forall f \in L^\infty(M), \quad \|f\|_{d_g} := \sup_{x \neq y} \frac{\lvert f(x)-f(y) \rvert}{d_g(x,y)},
$$
so that the *Lipschitz algebra* of the metric space $(M,d_g)$ is
$$
\operatorname{Lip}(M,d_g) := \{f \in L^\infty(M) \mid \|f\|_{d_g} < +\infty\} = \{f \in C(M) \mid \|f\|_{d_g} < +\infty\},
$$
which is a Banach algebra for the norm $\|\cdot\|_{\operatorname{Lip}(M,d_g)}$ defined by
$$
\forall f \in \operatorname{Lip}(M,d_g), \quad \|f\|_{\operatorname{Lip}(M,d_g)} := \|f\|_\infty + \|f\|_{d_g}.
$$
Proving the famous formula, then, boils down to proving the following two claims, where your Question 1 pertains to the first claim, whilst your Question 2 pertains to the second:

**Claim 1:** For all $x,y \in M$, $$d_g(x,y) = \sup\{\lvert f(x) - f(y) \rvert \mid f \in \operatorname{Lip}(M,d_g), \; \|f\|_{d_g} \leq 1\}.$$

**Claim 2:** Suppose that $M$ is spin$^\mathbb{C}$ and that $D$ is a
spin$^\mathbb{C}$ Dirac operator on a spinor bundle $S \to M$. Then,
for any $f \in L^\infty(M)$, $[D,f] \in B(L^2(M,S))$ if and only if $f
> \in \operatorname{Lip}(M,d_g)$, in which case $$\|[D,f]\| = \|f\|_{d_g}. $$

Let me first turn to your **Question 1**. ~~In principle, the algebra you're really taking a supremum over in Claim 1 is
$$
\{f \in L^\infty(M) \mid df \in L^\infty(M,T^\ast M)\} = \{f \in C(M) \mid df \in L^\infty(M,T^\ast M)\} = \operatorname{Lip}(M,d_g),
$$
where, for given $x$, $y \in M$, the supremum, which involves the Lipschitz seminorm $\|\cdot\|_{d_g}$, is attained by $d_g(x,\cdot) \in \operatorname{Lip}(M,d_g)$. Since $C^\infty(M)$ is ~~*not* dense in $\operatorname{Lip}(M,d_g)$ endowed with the Banach algebra norm induced by $\|\cdot\|_{d_g}$—indeed, the closure, from what I understand, will be the proper closed subalgebra $C^1(M)$ of $C^1$ functions on $M$—I suspect the answer is **no**, unless there's a result out there that lets you approximate Lipschitz functions by smooth functions in the Lipschitz seminorm alone (whilst necessarily doing violence to the uniform norms). **Edit:** As Nik Weaver points out in his comment, the answer is indeed **yes**. Fix $x$, $y \in M$. Let $r = d_g(x,\cdot)$, which is smooth on the complement of $\{x\} \cup \text{cut locus of $x$}$ and has Lipschitz constant $\|r\|_{d_g} = 1$. Then, by mollification on geodesic neighbourhoods of $x$ and of the cut locus of $x$ (cf. the approximation results of Greene and Wu, 1972, 1976, 1979), one can construct for any $\epsilon > 0$ a smooth function $f_\epsilon$ such that $\|r-f_\epsilon\|_\infty < \epsilon/2$ and $\|f_\epsilon\|_{d_g} \leq 1$, so that
$$
\lvert f_\epsilon(x) - f_\epsilon(y) \rvert = \lvert (r(x)-r(y)) + (r(x)-f_\epsilon(x)) + (r(y) - f_\epsilon(y)) \rvert \geq d_g(x,y) - \epsilon,
$$
and hence
$$
d_g(x,y) - \epsilon \leq \sup\{\lvert f(x) - f(y) \rvert \mid f \in C^\infty(M), \; \|f\|_{d_g} \leq 1 \} \leq d_g(x,y).
$$
Note that in context, in Connes's book and in the 1995 paper *Noncommutative geometry and reality*, $\mathcal{A}$ is either $L^\infty(M)$ or
$$
\{f \in L^\infty(M) \mid [D,f] \in B(L^2(M,S))\} = \operatorname{Lip}(M,d_g),
$$
where $D$ is the spin Dirac operator on the spinor bundle $S \to M$ on the spin manifold $M$.

Let me now turn to your **Question 2**. The essential point is that Claim 2 holds for *any* essentially self-adjoint *Dirac-type* operator on $M$, i.e., a first-order differential operator $D$ on a Hermitian vector bundle $E \to M$ such that
$$
D^2 = -g^{ij}\partial_i \partial_j + \text{lower order terms};
$$
spin$^\mathbb{C}$ and spin Dirac operators are certainly Dirac-type operators, but so is the Hodge–de Rham operator $d+d^\ast$ on $\wedge^\ast T^\ast M$, which only requires an orientation and a Riemannian metric on $M$. The reason is that an essentially self-adjoint Dirac-type operator $D$ *always* defines a self-adjoint Clifford action $c : T^\ast M \to \operatorname{End}(E)$ on $E$ by
$$
\forall f \in C^\infty(M), \quad c(df) := i[D,f],
$$
so that, in general, if $f \in L^\infty(M)$, then $[D,f] \in B(H)$ if and only if $f \in \operatorname{Lip}(M,d_g)$, with
$$
\|[D,f]\|_{B(H)} = \|c(df)\|_{B(H)} = \|c(df)^2\|^{1/2}_{B(H)} = \|g^{-1}(df,df)\|^{1/2}_\infty = \|f\|_{d_g}.
$$