We have such a definition:
Given an algebra $A$ over the field $K$, and $M$ is an $A$-bimodule.
Derivation is a linear map:
$$D:A \to M$$
That satisfies: $D(a*b)=D(a)\bullet b + a\bullet D(b)$
(With respect to given operations)
Then we define tangent vector $X$ on manifold $M$ at point $p$ as a map: $$X:C^{\infty}(M)\to \mathbb{R}$$ (Where $C^{\infty}(M)$ is a $\mathbb{R}$-algebra of smooth functions on M)
Question: Implication that tanget vector is a derivation is correct, if $\mathbb{R}$ is a $C^{\infty}(M)$-bimodule.
But how do we see that? Since $\mathbb{R}$ is a field $\Rightarrow$ ring $\Rightarrow$ $\mathbb{R}$-$\mathbb{R}$-bimodule over itself. But what about algebra of smooth functions?
You can interpret a tangent vector at $p \in M$ as a derivation in your sense. A tangent vector at $p \in M$ is a map $v \colon C^{\infty}(M) \rightarrow \mathbb{R}$ which is $\mathbb{R}$-linear and satisfies
$$ v(fg) = f(p)v(g) + v(f)g(p). $$
If you consider $\mathbb{R}$ as $C^{\infty}(M)$-bimodule by the structure maps
$$ f \bullet x = f(p) \cdot x, \,\,\, x \bullet g = g(p) \cdot x $$
where $\cdot$ is the regular multiplication of real numbers then the equation above becomes
$$ v(fg) = f \bullet v(g) + v(f) \bullet g. $$
In more fancy terms, you have the evaluation homomorphism $\operatorname{ev}_p \colon C^{\infty}(M) \rightarrow \mathbb{R}$ and $\mathbb{R}$ is a bimodule over itself so by pulling back using $\operatorname{ev}_p$ we give $\mathbb{R}$ a bimodule structure over $C^{\infty}(M)$.