Directional Derivative of functions $f:S\rightarrow\mathbb{R}$ Suppose $S\subseteq\mathbb{R}^3$ is a regular surface and let $f:S\rightarrow \mathbb{R}$ is differentiable. I know that a vector $v\in T_pS$ is of the form $v=\alpha'(0)$, where $\alpha:]-\epsilon,\epsilon[\rightarrow S$ is such that $\alpha(0)=p$. Now we have defined the directional derivative of f with direction v at point p as $Df_p(v)=(f\circ\alpha)'(0)$.
My goal is to prove that $Df_p$ is linear. And this should be easy but I don't know how to it. 
Let $v,w\in T_pS$ and choose $\alpha,\beta$ curves such that $\alpha'(0)=v$, $\beta'(0)=w$. I want to prove that $D_{v+w}f(p)=D_vf(p)+D_wf(p)$, right? Should I fix $\gamma$ such that $\gamma'(0)=v+w$? I don't seem to go anywhere.
Thank you very much!

Yes, this sort of abstract notation can be confusing sometimes. Let me write your expression for the directional derivative in coordinates, expanding it out using the chain rule:
$$ Df_p (v) = \frac{d(f \circ \alpha)} {dt} (0)= \frac{\partial f}{\partial x}(p) \frac{d \alpha_x}{dt}(0) + \frac{\partial f}{\partial y}(p) \frac{d \alpha_y}{dt}(0) + \frac{\partial f}{\partial z}(p) \frac{d \alpha_z}{dt}(0) $$
$$ \ \ \ \ = \frac{\partial f}{\partial x}(p) v_x + \frac{\partial f}{\partial y}(p) v_y + \frac{\partial f}{\partial z}(p)v_z.$$
Similarly, replacing $v$ with $w$ and replacing $\alpha$ with $\beta$, we get
$$ Df_p (w) = \frac{\partial f}{\partial x}(p) w_x + \frac{\partial f}{\partial y}(p) w_y + \frac{\partial f}{\partial z}(p)w_z.$$
Finally, replacing $v$ with $v+ w$ and replacing $\alpha$ with $\gamma$, we get
$$ Df_p (v+w) = \frac{\partial f}{\partial x}(p) (v_x+ w_x) + \frac{\partial f}{\partial y}(p) (v_y + w_y) + \frac{\partial f}{\partial z}(p) (v_z + w_z).$$
Now that we have written all of this out, it should be clear that
$$ Df_p (v+w) = Df_p(v) + Df_p(w).$$
which is the same as saying that $Df_p$ acts linearly on tangent vectors in $T_p$.
