"Pedantic" derivation of geodesic equation using pullback bundles I'm trying to get more comfortable with manipulations involving connections and vector fields so I've tried to derive the geodesic equations without having to resort to any familiarities using standard calculus, everything computed "properly" from the definitions.
For a Reimannian manifold $(M,g)$ I have a curve $\gamma : \mathbb{R} \rightarrow M$ which in local coordinates can be written as $\gamma(t) = \left(x^1(t), \ldots, x^n(t)\right)$. If I wish $\gamma(t)$ to be a geodesic then I want its tangent vector to be auto-parallel.
The tangent vector is given by the pushforward of coordinate vector field on $\mathbb{R}$ $``\hspace{01mm}\dot{\gamma}(t)\hspace{-0.5mm}" := \gamma_*\left(\frac{\partial}{\partial t}\right) = \dot{x}^i(t) \frac{\partial}{\partial x^i}$
I want $\nabla_\dot\gamma \dot{\gamma} = 0$, but this expression is misleading since the vector field $\dot{\gamma}(t)$ only exists along the image of $\gamma(t)$, but we can consider the pullback of $M$ by $\gamma$ and take the connection and vector bundle with us. If $\nabla$ is the Levi-Civita connection on $(M,g)$ denote $\widetilde{\nabla}$ as its pullback connection by $\gamma$
Then $\gamma$ is geodesic if $\widetilde{\nabla}_\frac{\partial}{\partial t}\gamma_*\left(\frac{\partial}{\partial t}\right) = 0$. At this point I start to get stuck, I have the following definition from a worked exam question that inspired me to do this exercise:

I'm not quite sure if I'm in the lucky situation where $\gamma_*\left(\frac{\partial}{\partial t}\right)$ is already of the form $v \circ u$, and I'm not so sure what this even means, if my bundle is the tangent bundle of $M$, then my sections $e_i = \frac{\partial}{\partial x^i}$ are vector fields, how does one compose a vector field with a map?
I think this has something to do with where we are evaluating $\gamma_*\left(\frac{\partial}{\partial t}\right)f = \left.\dot{x}^i(t) \frac{\partial f}{\partial x^i}\right|_{\gamma(t)}$, that is, $e_i = \left.\frac{\partial}{\partial x^i}\right|_p$ for $p \in M$ whereas $e_i \circ \gamma = \left.\frac{\partial}{\partial x^i}\right|_{\gamma(t)}$. I'm not sure how to properly justify this but it certainly feels more correct that $\gamma_*\left(\frac{\partial}{\partial t}\right)$ should be "evaluating" on $\gamma(t)$ rather than any old $p$ since the whole point of this pullback stuff was to differentiate along the curve.
If we accept the above handwaving then my calculation is as follows:
$$\widetilde{\nabla}_\frac{\partial}{\partial t}\gamma_*\left(\frac{\partial}{\partial t}\right) := \nabla_{\gamma_*\left(\frac{\partial}{\partial t}\right)}\gamma_*\left(\frac{\partial}{\partial t}\right) 
= \nabla_{\dot{x}^i(t) \frac{\partial}{\partial x^i}}\dot{x}^j(t) \frac{\partial}{\partial x^j}$$
Using $C^\infty(M)$ linearity of a connection in the lower argument and the Liebnitz rule gives
$$ = \dot{x}^i(t) \nabla_{\frac{\partial}{\partial x^i}}\dot{x}^j(t) \frac{\partial}{\partial x^j} = \dot{x}^i(t)\frac{\partial}{\partial x^i}\left(\dot{x}^j\right)\frac{\partial}{\partial x^j} +  \dot{x}^i(t)\dot{x}^j(t)\nabla_{\frac{\partial}{\partial x^i}}\frac{\partial}{\partial x^j}$$
The second term is $\dot{x}^i(t)\dot{x}^j(t)\Gamma_{ij}^k\frac{\partial}{\partial x^k}$ which starts to look on the right tracks, but I have no idea what to do to the first term to get a second time derivative, and if my approach is even correct.
Apologies for the wall of equations, but I wanted to get down all my thoughts and where my confusions lie, I am looking for how to finish the derivation and an explanation of all this stuff with pullback bundles and correct any misunderstandings I have. Thanks in advance.
 A: This is a good question. Here's how to do it: given coordinates $(x^1,\ldots, x^n)$ around (some particular point) $\gamma(t)$, where $\gamma\colon I \to M$, we have that $$\dot{\gamma}(t) = \sum_{i=1}^n \dot{x}^i(t)\frac{\partial}{\partial x^i}\bigg|_{\gamma(t)}$$for all $t \in I$. Then indeed $\nabla_{\dot{\gamma}(t)}\dot{\gamma}$ does not immediately makes sense, but we have the pull-back bundle $\gamma^*(TM) \to I$, and a connection $\gamma^*\nabla$. Then $\partial/\partial t$ is a vector field on the "base manifold" $I$, and $\gamma_\ast(\partial/\partial t) = \dot{\gamma}$, and this is what allows us to use the defining property of $\gamma^*\nabla$: $$\begin{align}\frac{D\gamma'}{{\rm d}t}(t) &= (\gamma^*\nabla)_{(\partial/\partial t)|_t}(\dot{\gamma}) = (\gamma^*\nabla)_{(\partial/\partial t)|_t}\left(\sum_{j=1}^n \dot{x}^j \left(\frac{\partial}{\partial x^j}\circ \gamma\right)\right) \\ &= \sum_{j=1}^n \ddot{x}^j(t) \frac{\partial}{\partial x^j}\bigg|_{\gamma(t)}+ \sum_{j=1}^n \dot{x}^j(t) (\gamma^*\nabla)_{(\partial/\partial t)|_t}\left(\frac{\partial}{\partial x^j}\circ \gamma\right) \\ &= \sum_{k=1}^n \ddot{x}^k(t) \frac{\partial}{\partial x^k}\bigg|_{\gamma(t)}  + \sum_{j=1}^n \dot{x}^j(t) \nabla_{\dot{\gamma}(t)}\frac{\partial}{\partial x^j} \\ &= \sum_{k=1}^n \ddot{x}^k(t) \frac{\partial}{\partial x^k}\bigg|_{\gamma(t)}  + \sum_{j=1}^n \dot{x}^j(t) \nabla_{\sum_{i=1}^n \dot{x}^i(t) (\partial/\partial x^i)|_{\gamma(t)}}\frac{\partial}{\partial x^j} \\ &= \sum_{k=1}^n \ddot{x}^k(t) \frac{\partial}{\partial x^k}\bigg|_{\gamma(t)}  + \sum_{i,j=1}^n \dot{x}^i(t)\dot{x}^j(t) \nabla_{(\partial/\partial x^i)|_{\gamma(t)}}\frac{\partial}{\partial x^j} \\ &= \sum_{k=1}^n \ddot{x}^k(t) \frac{\partial}{\partial x^k}\bigg|_{\gamma(t)}  + \sum_{i,j,k=1}^n \Gamma_{ij}^k(\gamma(t))\dot{x}^i(t)\dot{x}^j(t) \frac{\partial}{\partial x^k}\bigg|_{\gamma(t)} \\ &= \sum_{k=1}^n \left(\ddot{x}^k(t) + \sum_{i,j=1}^n \Gamma_{ij}^k(\gamma(t))\dot{x}^i(t)\dot{x}^j(t)\right)\frac{\partial}{\partial x^k}\bigg|_{\gamma(t)}.\end{align}$$
