basic definition of vector-fields on a manifold Many textbooks introduce vector-fields on a manifold $M$ along the lines of $ X = X_i \frac{\partial}{\partial x_i} $ where -without further ado- $\frac{\partial}{\partial x_i}$ is introduced as a basis vector in Tangent space. Given that the same symbol is widely used as a differential operator in mathematics, should I think about $X$ in terms of an operator rather than a vector?  Or is that notation a clever suggestive trick to help ease calculations - similar to physicists speaking of "ket-" $ | \psi>$ and "bra-" $ < \Phi | $ vectors in a physical Hilbert space that "magically" turn into a scalar-product $ <\Phi|\Psi> $ when "meeting" in a calculation? Similar conceptual difficulties arise for me when in the the definition of 1-forms $\omega = \omega_1 dx $ the object $dx$ is introduced as a "unit" vector when I am used to thinking about $dx$ as an infinitesimal quantity. Cheers!
 A: Since no one answered this properly, I'll do it for completeness. Yes, you can regard $X$ as an operator, taking a function $f$ to a function $X(f)$, $p \mapsto X_p(f) = {\rm d}f_p(X_p)$.
Given a coordinate system $\varphi = (x^i)$ for your manifold, we set $$\frac{\partial f}{\partial x^i}(p) \doteq \frac{\partial (f \circ \varphi^{-1})}{\partial x^i}(\varphi(p)),$$where the derivative in the right side is the ordinary partial derivative in Euclidean space. It does not make sense to talk about partial derivatives in an arbitrary manifold without making a choice of coordinate system to transfer the coordinate axes in $\Bbb R^n$ to "coordinate axes" in a manifold. That being said, people denote $\partial/\partial x^i$ by $\partial_i$ to simplify the notation. We define the action of $X = X^i\partial_i$ on a function $f$ by $X(f) = X^i\partial_if$. Indeed, if $(\widetilde{x}^i)$ is another coordinate system in the manifold, and we write $X = \widetilde{X}^i\widetilde{\partial_i}$, then the chain rule gives $$\widetilde{X}^i\widetilde{\partial_i}f = \left(X^j \frac{\partial \widetilde{x}^i}{\partial x^j} \right)\left(\frac{\partial x^k}{\partial \widetilde{x}^i} \partial_kf\right) = X^j \delta^k_j \partial_kf = X^j\partial_jf.$$
