Covariant derivative and geodesic Let $f: U \subset \mathbb{R}^2 \rightarrow \mathbb{R}^3$ be a surface patch. Then if we have two vector fields $$X = \sum_i \xi^i \frac{\partial f}{\partial u^i}$$
and $$Y = \sum_i \eta^i \frac{\partial f}{\partial u^i}$$
the covariant derivative is given by $$\nabla_X Y:= \sum_{i,k} \xi^{i} \left( \frac{\partial \eta^k}{\partial u^i} + \sum_j \Gamma_{i,j}^k \eta^j\right)\frac{\partial f}{\partial u^k}$$
Now my question is: If I have a curve $\gamma: I \rightarrow U$ and $c:=f \circ \gamma,$
then my textbook claims that 
$$\nabla_{c'}c' = \sum_{i,k} \dot{u}^{i}(t) \left( \frac{\partial \dot{u}^{k}(t)}{\partial u^i} + \sum_j \Gamma_{i,j}^k \dot{u}^{j}(t)\right)\frac{\partial f}{\partial u^k}$$
Although I don't know what this function $u$ should actually denote, this equation looks somehow wrong, as $\dot{u}$ is a function of $t$ but is differentiated with respect to $u^{i}$, which does not make sense to me. Could anybody explain this to me?
So I have essentially two questions:
1.)  How is $u$ defined?
2.) How does this equation follow from the definition of the covariant derivative and why is the derivative of $\dot{u}$ with respect to $u^{i}$ not zero?
 A: Observe that through the map 
$$\left(\begin{array}{c}
u^1\\
u^2
\end{array}\right)
\stackrel{f}\to
\left(\begin{array}{c}
x^1\\
x^2\\
x^3
\end{array}\right),
$$
you have a GPS for points in the surface. It is with this that you get $\frac{\partial f}{\partial u^1}$ and $\frac{\partial f}{\partial u^2}$, which are the two columns of the jacobian $Jf$, and serve to define the tangent space at any point on the surface.
So to travel along a curve in the surface you need to travel along a curve in the domain $U$. This implies that $u^1,u^2$ depends on a parameter, which we can denote with $t$.
Schematically 
$$t\stackrel{\gamma}\to
\left(\begin{array}{c}
u^1(t)\\
u^2(t)
\end{array}\right)
\stackrel{f}\to
\left(\begin{array}{c}
x^1(t)\\
x^2(t)\\
x^3(t)
\end{array}\right),
$$
will give the way to control the positions on curve over the surface.
Now $\dot{c}=\frac{d}{dt}(f\circ\gamma)=Jf\cdot\dot{\gamma}$ and since $\dot{\gamma}=\dot{u^1}e_1+\dot{u^2}e_2$ then 
$$\dot{c}=Jf(\dot{u^1}e_1+\dot{u^2}e_2)=\dot{u^1}Jf(e_1)+\dot{u^2}Jf(e_2),$$
so
$$\dot{c}=\dot{u^1}\frac{\partial f}{\partial u^1}+\dot{u^2}\frac{\partial f}{\partial u^2}.$$
This way you can use it in $\nabla_{\dot{c}}\dot{c}$ to get 
$$\nabla_{\dot{c}}\dot{c}=\sum_s(\ddot{u^s}+\sum_{ij}\Gamma^s{}_{ij}\dot{u^i}\dot{u^j})\frac{\partial f}{\partial u_s}.$$

This last formula is get as follow:
$$\nabla_{\dot{c}}\dot{c}=
\nabla_{\sum_s\dot{u^s}\frac{\partial f}{\partial u_s}}
\sum_{\sigma}\dot{u^{\sigma}}\frac{\partial f}{\partial u_{\sigma}},$$
$$\quad=\sum_{s{\sigma}}\dot{u^s}\nabla_{\frac{\partial f}{\partial u_s}}
\dot{u^{\sigma}}\frac{\partial f}{\partial u_{\sigma}},$$
$$\qquad=\sum_{s{\sigma}}\dot{u^s}
\frac{\partial \dot{u^{\sigma}}}{\partial u_s}\frac{\partial f}{\partial u_{\sigma}}+\sum_{s{\sigma}}\dot{u^s}\dot{u^{\sigma}}\nabla_{\frac{\partial f}{\partial u_s}}\frac{\partial f}{\partial u_{\sigma}},$$
$$\qquad=\sum_{s{\sigma}}\dot{u^s}
\frac{\partial \dot{u^{\sigma}}}{\partial u_s}\frac{\partial f}{\partial u_{\sigma}}+\sum_{s{\sigma}r}\dot{u^s}\dot{u^{\sigma}}\Gamma^r{}_{s{\sigma}}\frac{\partial f}{\partial u_r},$$
$$\qquad=\sum_{s{\sigma}}
\frac{\partial \dot{u^{\sigma}}}{\partial u_s}\frac{du^s}{dt}\frac{\partial f}{\partial u_{\sigma}}+\sum_{s{\sigma}r}\dot{u^s}\dot{u^{\sigma}}\Gamma^r{}_{s\sigma}\frac{\partial f}{\partial u_r},$$
$$\qquad=\sum_{{\sigma}}
\frac{d^2u^s}{dt^2}\frac{\partial f}{\partial u_{\sigma}}+\sum_{s\sigma r}\dot{u^s}\dot{u^r}\Gamma^{\sigma}{}_{sr}\frac{\partial f}{\partial u_{\sigma}},$$
$$\qquad=\sum_{{\sigma}}\left(
\frac{d^2u^{\sigma}}{dt^2}+\sum_{sr}\dot{u^s}\dot{u^r}\Gamma^{\sigma}{}_{sr}\right)
\frac{\partial f}{\partial u_{\sigma}}.$$
