TITLE: Vector fields on manifolds
QUESTION [9 upvotes]: I have recently started a course on differential geometry (from a physicists perspective) and I am having trouble convincing myself why vector fields are represented as differential operators on manifolds?
Is it because, in general, one cannot define a vector on a manifold as an arrow between two points on the manifold (as it generally will not be Euclidean) and as such the best we can do is consider the tangent space at each point on the manifold. Now, we could define vectors as the directional derivatives of the coordinate curves at that point, however this is very much coordinate dependent, which doesn't fit in with the notion of a vector. As such, if we consider tangent vector at a point as a differential operator that defines a tangent space at that point, then we satisfy the axioms of a vector space and thus achieve the notion of vectors on manifolds?
Apologies for any inaccuracies and the wordiness of my attempt to answer my problem. I would really appreciate an intuitive explanation if possible.
Thanks for your time.

REPLY [4 votes]: I think your current intuition is pretty good, let me see if I can make it a bit more precise while still keeping things conceptual.  I'm not sure how you defined tangent vectors in your class but one way of doing this is as follows. 
Pick a point $x \in M$.  A tangent vector at $x$ is an equivalence class of curves $\gamma : (-1,1) \to M$ satisfying $\gamma(0) = x$ under the equivalence relation that they are tangent at $x$.  The definition of "tangent at $x$" actually requires one to pick a chart on a neighborhood of $x$ and then transform into the statement that the gradients at $0$ are equal in $R^n$; it turns out that this is chart independent and give a real equivalence relation.  So a tangent vector really does feel like an arrow in $M$ with this definition.  Also interpreting a vector field as a function taking each $x \in M$ to a tangent vector at $x$ seems pretty reasonable.  Recall we refer to the set of all tangent vectors at $x \in M$ as the tangent space $T_xM$ and the collection of all tangent spaces form a new manifold $TM$ called the tangent bundle (lots of grunt work goes into proving this).
Out of this nice geometric picture you can also see how to define a "differential operator" from a vector field $v$.  Work pointwise in $M$.  Given any differentiable function $f : M \to R$ you can operate on the function $f$ with the tangent vector in $v(x) \in T_xM$ by taking a representative curve $\gamma : (-1,1) \to M$, making the composition $f \circ \gamma : (-1,1) \to R$ and taking the derivative at zero $(f \circ \gamma)^\prime(0)$ (this gives you a number).  Now one has to show that this depends only on the equivalence class of $\gamma$ (i.e. depends only on $v(x)$) and in fact defines a smooth function $M \to R$ as $x$ varies over $M$.  One can work in charts to validate that the operation we've defined satisfies the product rule etc.  So in this way the vector field $v$ give a differential operator.  Note that all of this boils down to looking at how a single tangent vector allows us to assign a number to a function; this way of thinking of a tangent vector is usually described by calling it a derivation.
It turns out you can start just the idea of differentiation/derivations and define your tangent space from that (which is what I am guessing your class did).  Doing so has certain advantages such as avoiding charts and equivalence classes.  Also, if the notion of a derivation is interpreted appropriately the construction via differentiation has a certain generality because it is purely "algebraic".
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