Geodesic equation along curves Let $M$ be a smooth manifold and $X$ a vector fields on $M$. Let $\{x_i\}_{i=1}^n$ be local coordinates, so $X=X^i \partial_i$
The equation for auto parallel transport (geodesic) of $X$ is, in coordinates,
$$(\nabla_X X)^i= X^k(\partial_k X^i + \Gamma^{i}_{\,jk}X^j)=0. $$
So far so good. The problem comes when $X$ is defined to be a vector field along a curve. Let $\dot{\gamma}(t)$ be the velocity vector of the curve $\gamma$, in principle nothing should change, and simply $X^k=(x^k\circ \gamma)'(t)$, the problem is that above $X$ was a vector field on the manifold, so $X^k\in C^\infty(M)$, but now $X^k: \mathbb{R}\rightarrow\mathbb{R}$, $$X^k = \frac{\mathrm{d}}{\mathrm{d}t}\gamma^k$$ where $\gamma^k=x^k\circ \gamma$. Naïvely substituting term by term I get 
$$\frac{\mathrm{d}\gamma^k}{\mathrm{d}t}\partial_k \frac{\mathrm{d}\gamma^i}{\mathrm{d}t} =-\frac{\mathrm{d}\gamma^k}{\mathrm{d}t} \Gamma^{i}_{\,jk}\frac{\mathrm{d}\gamma^j}{\mathrm{d}t} $$
This doesn't make much sense to me, the problem is I don't know how to interpret $$\partial_k \frac{\mathrm{d}\gamma^i}{\mathrm{d}t}$$
As $\partial_k$ is a map on $C^\infty(M)$ and $\frac{\mathrm{d}\gamma^i}{\mathrm{d}t}:\mathbb{R}\rightarrow\mathbb{R}$. I don't know if I can exchange the two derivatives and interpret it as
$$ \frac{\mathrm{d}}{\mathrm{d}t}(\partial_k x^i)(\gamma(t))=\frac{\mathrm{d}\gamma^k}{\mathrm{d}t}$$
but even if I did I would end up with a derivative squared, not a second derivative. Basically I'm lost and I feel that I seriously misunderstood something: I think the core problem is my definition of a vector field on a manifold and one along a curve don't agree. 
How can I solve this problem? I want to manipulate the first equation to get the geodesic equation 
$$\frac{\mathrm{d}^2\gamma^i}{\mathrm{d}t^2}+\Gamma^i_{\,jk}\frac{\mathrm{d}\gamma^k}{\mathrm{d}t}\frac{\mathrm{d}\gamma^j}{\mathrm{d}t}=0 $$
 A: Before finding the geodesic equation, you should be comfortable with the covariant derivative of a vector field along a curve. So let $\gamma=(\gamma^1,\ldots,\gamma^n)$ be a curve, and let $X(t)=X^i(t)\partial_i$ be a vector field along $\gamma$. You are right, to begin with; as $X$ is only defined along $\gamma$, there isn't much sense in expressions like $\partial_kX^i$. The way to solve this is first compute the covariant derivative of $\partial_i$ (which is defined on a coordinate neighborhood) and then use the Leibniz rule. So we start with
$$\frac{D}{dt}\partial_i=\nabla_{\dot{\gamma}}\partial_i=\dot{\gamma}^j\Gamma_{ji}^k\partial_k.$$
Then, by forcing the Leibniz rule to hold, 
$$\begin{align}\frac{D}{dt}X&=\frac{D}{dt}X^i\partial_i\\&=\dot{X}^i\partial_i+X^i\dot{\gamma}^j\Gamma_{ji}^k\partial_k\\&=\left(\dot{X}^k+X^i\dot{\gamma}^j\Gamma_{ji}^k\right)\partial_k.\end{align}$$
Substituting $X=\dot{\gamma}$,
$$\frac{D}{dt}\dot{\gamma}=\left(\ddot{\gamma}^k+\dot{\gamma}^i\dot{\gamma}^j\Gamma_{ji}^k\right)\partial_k.$$
