Here is a counterexample:

Let $M$ be the one-dimensional unit circle, so we can identify $C^1(M)$ with \begin{align*} X = \{f \in C^1([0,1]): \, f(0) = f(1), \; f'(0) = f'(1)\}. \end{align*}

For each $z \in [0,1)$ let $d_z \in X^*$ be given by $\langle d_z,f\rangle = f'(z)$ for each $f \in X$. Then the set \begin{align*} K := \{d_z: \, z \in [0,1)\} \end{align*} is weak${}^*$-compact (since $M$ is compact), but $K$ cannot be separated from $0$ by an element of $X$. Indeed, let $f \in X$. If $f$ is constant, then every functional in $K$ vanishes on $f$. If $f$ is not constant then, due to the periodic nature of $f$, there exist $z_1,z_2 \in [0,1)$ such that $f'(z_1) < 0$ and $f'(z_2) > 0$. Hence, a convex combination of $d_{z_1}$ and $d_{z_2}$ vanishes on $f$.

Remark. Funnily enough, this counterexample uses that the manifold $M$ is closed. I don't know how to construct a similar counterexample on $C^1([0,1])$.

Here is a counterexample:

Let $M$ be the one-dimensional unit circle, so we can identify $C^1(M)$ with \begin{align*} X = \{f \in C^1([0,1]): \, f(0) = f(1), \; f'(0) = f'(1)\}. \end{align*}

For each $z \in [0,1)$ let $d_z \in X^*$ be given by $\langle d_z,f\rangle = f'(z)$ for each $f \in X$. Then the set \begin{align*} K := \{d_z: \, z \in [0,1)\} \end{align*} is weak${}^*$-compact (since $M$ is compact), but $K$ cannot be separated from $0$ by an element of $X$. Indeed, let $f \in X$. If $f$ is constant, then every functional in $K$ vanishes on $f$. If $f$ is not constant then, due to the periodic nature of $f$, there exist $z_1,z_2 \in [0,1)$ such that $f'(z_1) < 0$ and $f'(z_2) > 0$. Hence, a convex combination of $d_{z_1}$ and $d_{z_2}$ vanishes on $f$.

Remark (Edited after a comment by Tom Goodwillie). The counterexample above uses that the manifold $M$ is closed. The comment by Tim Goodwillie below explains how this example can be adjusted to obtain a counterexample on $C^1([0,1])$.