When vectors act on scalars. 
Background. I've been struggling through an introduction to differential geometry this semester. Recently, a tiny part of what we've been learning "clicked" for me, and to solidify this, I'd like to get some further information and especially just understand the relevant terminology.
Hence this question.

Fix a smooth manifold $M$. By a scalarfield on $M$, I mean a smooth function $M \rightarrow \mathbb{R}$. Write $S$ for the ring of scalarfields on $M$. By a vectorfield on $M$, I mean a derivation on $S$; explicitly, this is an $\mathbb{R}$-linear function $v : S \rightarrow S$ satisfying the Leibniz product law: $$v(s_0 s_1) = v(s_0)s_1 + s_0 v(s_1)$$
Write $V$ for the collection of vectorfields on $M$.
There's something odd about this situation. In the basic commutative algebra I'm familiar with, "scalars" (i.e. the elements of some commutative ring) act on "vectors" (i.e. the elements of some abelian group.) That is certainly the case here; we have a multilinear function: $$S, V \rightarrow V$$
that satisfies $s_1(s_0v) = (s_1s_0)v$ and $1_S v = v$, hence $V$ is an $S$-module.
However we can also go the other way; vectorfields can act on scalarfields by $v,s \mapsto v(s)$. This gives a multilinear function: $$V,S \rightarrow S$$
It satisfies $v(s_0s_1) = v(s_0)s_1+s_0 v(s_1)$.
Furthermore, these two actions are related by: $s_0 (vs_1) = (s_0 v)s_1$
Furthermore, given $s \in S$, we can define the differential $ds : V \rightarrow S$ by writing $ds(v) = v(s)$. Hence $ds$ is an element of the dual space $V^*,$ where $V$ is viewed as an $S$-module.

Question. What terminology surrounds this situation? For example:
  
  
*
  
*What do we call multilinear maps $V,S \rightarrow S$ that satisfy the Leibniz law?
  
*What kind of a structure is formed by the whole data bundle consisting of $S$ and $V$, together with the actions $S,V \rightarrow V$ and $V,S \rightarrow S$?
  
*Are other there any basic relationships here that I really need to be aware of? If I understand correctly, $V$ forms a "Lie algebra", but I'm not sure of the relevance of this.
  

 A: A linear map $S\to S$ satisfying the Leibniz rule is called a derivation.
It seems that you already know a lot. Maybe now you should get familiar with the language of vector bundles. For example, a smooth function $M\to\mathbb{R}$ can be thought of as a section of the trivial bundle $M\times\mathbb{R}$. The tangent bundle $TM$ consists of all tangent spaces at all the points of $M$. Hence, a vector field is a section of the tangent bundle. The cotangent bundle $T^*M$ is the dual to the tangent bundle. Hence, the differential of a function is a section of $T^*M$. In general, a section of $T^*M$ is called a differential $1$-form. By taking tensor products of the tangent and cotangent bundles, we obtain many other vector bundles. Any section of such a bundle is called a tensor field.
Yes, the collection of vector fields forms a Lie algebra. The most basic meaning of this statement is the existence of the operator $[\cdot,\cdot]$ which eats two vector fields and spits out another vector field. This operator is uniquely defined by the property$$[X,Y](f)=X(Y(f))-Y(X(f)),\qquad X,Y\in V,f\in S.$$
