From http://en.wikipedia.org/wiki/Tangent_space#Definition_via_derivations, I understand that if $\gamma: (-1,1) \to M$ is a curve (and $M$ a manifold), with tangent vector $\gamma'(0)$, then the corresponding derivation is $D_{\gamma}(f) = \frac{d}{dt}\mid_{t=0} f(\gamma(t)) = (f \circ \gamma)'(0)$.
I read that $\{ \frac{\partial}{\partial x_i}\mid_p \}$ is a basis for $T_pM$, the tangent space at point $p \in M$, given a coordinate chart $(x,U)$. And that by definition, $\frac{\partial}{\partial x_i}\mid_p(f) = D_i(f\circ x^{-1})(x(p))$, where $D_i$ is the ordinary $i$-th partial derivative of a function from $\mathbb{R}^n$ to $\mathbb{R}$.
What I am confused about is this: using the definition in the first paragraph, I would expect $\frac{\partial}{\partial x_i}\mid_p(f) = (f \circ \gamma)'(0)$ for some curve $\gamma$. But somehow, we managed to get the ordinary partial derivative $D_i \mid_{x(p)}$ in there. How do I reconcile the two definitions?
Well, you should start like this, take a smooth function $f\in C^{\infty}(M)$, anda take a point $p\in M$ and vector in $\frac{\partial}{ \partial x_i}|_p\in T_pM$, than take a curve on $\gamma:(-\epsilon,\epsilon)\to M$ such that $\gamma(0)=p$ and $\gamma'(0)=\frac{\partial}{ \partial x_i}|_p$. Then you have $\frac{\partial}{ \partial x_i}|_pf=\gamma'(0) f= (d\gamma\frac{\partial}{ \partial t}|_{t=0}) f=\frac{d}{dt}|_{t=0}(f\circ \gamma)(t) =(f\circ\gamma)'(0)$. So this is a way to make the calculus easier.