Differential Geometry Generalized Functions A vector in differential geometry becomes a functional on the manifold $M$, namely
$X_{p,\gamma}:C^{\infty}(M)\to \mathbb{R}$,
$X_{p,\gamma}f=\left.\dfrac{d}{dt}\right|_{t=0}(f\circ \gamma)$. 
This reminds me a lot of what a generalized function is. Namely a generalized function "eats" up a smooth test function (that vanishes outside a bounded subset of $M$ etc etc) and "spits" out a number, however, symbolically like
$\left< f, \cdot \right>: C^{\infty}(U)\to \mathbb{R}$,
$\left< f,\phi \right>=\int f(x)\phi(x)dx$.
It feels that there must be some sort of connection here between these two concepts, but I am totally missing the details. (or the point!)
 A: Your question seems to be a bit difficult to answer to me. Formally, the answer certainly is yes, directional derivatives in a point do define distributions. Still, in my opinion this is a rather unnatural way to view this operation. Just think about the case of $\mathbb R^n$. You can certainly interpret a point $x\in\mathbb R^n$ and a vector $v\in\mathbb R^n$ as a generalized function, since $f\mapsto Df(x)(v)$ is a linear map. This is the directional derivative of the delta-distribution at $x$ in direction $v$, which of course is an very important concept. I dont think that you would view it as a way to understand directional derivatives of smooth functions. 
Here the situation is similar. The curve $\gamma$ is used to define a direction at a point, namely the direction $\gamma'(0)$ in the point $\gamma(0)$. Then you can differentiate smooth functions in that direction and you gave the the defining formula for this, which simply reflects the fact that the chain rule should be applicable to $f\circ\gamma$. You can interpret this as defining a generalized function (the directional derivative of $\delta_{\gamma(0)}$ in direction $\gamma'(0)$) but this seems to be a rather roundabout way to view things. 
