Let $X$ be a set and $\mathcal{F}\subseteq {\mathbb{R}^X}$ an arbitrary family of functions.
The **superposition closure** of $\mathcal{F}$ is defined as
$$
\overline{\mathcal{F}}=\{ H\circ(f_1\times\cdots f_i)\mid H\in C^\infty(\mathbb{R}^i),\ f_1,\ldots,f_i\in \mathcal{F},\ i\in\mathbb{N}\}\, ,
$$
where $(f_1\times\cdots f_i)(x)=(f_1(x),\ldots, f_i(x))$, $\forall x\in X$,
and it is a commutative unitary sub-algebra of ${\mathbb{R}^X}$ containing $\mathcal{F}$. For instance, if $X=\mathbb{R}^2$ and $\mathcal{F}=C^\infty(\mathbb{R})\otimes_{\mathbb{R}} C^\infty(\mathbb{R})$, then $\overline{\mathcal{F}}=C^\infty(\mathbb{R}^2)$ (basically, this is the Kolmogorov superposition theorem discussed in the post Kolmogorov superposition for smooth functions).

Suppose now that $\mathcal{F}$ is already an unitary sub-algebra of ${\mathbb{R}^X}$. Then, we have a sequence of algebra inclusions $$ {\mathcal{F}}\subseteq \overline{\mathcal{F}}\subseteq \mathbb{R}^X $$

QUESTION: is it true that any derivation of ${\mathcal{F}}$ can be uniquely extended to a derivation of $\overline{\mathcal{F}}$?

By **derivation** I simply mean an $\mathbb{R}$-linear map $\Delta:{\mathcal{F}}\to {\mathcal{F}}$ satisfying the Leibniz rule.

I've been trying for many years to answer this question: all the examples I could think of gave a positive answer, but I could not lay down a general proof. All the experts I contacted gave me unsatisfactory answers (basically, one uses the chain rule for derivatives and tries to prove that the result is independent on the choice of the $H$ and the $f$'s above). Finding a counterexample would be really nice!

EDITED AFTER M. BACHTOLD'S ANSWER:

The counterexample proposed in the answer below is correct, and based on the fact that there can be a subalgebra $\mathcal{F}$ of $C^\infty(\mathbb{R})$ whose (real) spectrum is $\mathbb{R}^2$: it suffices to take two algebraically independent smooth functions. Then the inclusion $\mathcal{F}\subseteq C^\infty(\mathbb{R})$ should correspond, at the level of spectra, to a map $\mathbb{R}\to \mathbb{R}^2$, i.e., a "curve". I guess that, in this case, the ambiguity boils down to the fact that there are many ways to extend a vector field on a curve to the ambient space, loosely speaking.
What I'm really looking for is a counterexample where $\mathcal{F}$ and its superposition closure $\overline{\mathcal{F}}$ **share the same spectrum**.