Changing Coefficients of Covariate derivative Given an affine connection $\nabla_X$ on some vector field $X,Y \in \mathcal{V}(M)$ I want to compute its form with local coefficients $\Gamma^k_{ij}$. I.e.
$$
\nabla_X(Y)|_U = \sum_{i=1}^{n} a_i \nabla_{\frac{\partial}{\partial x_i}}(\sum_{j=1}^{n}b_j \frac{\partial}{\partial x_j}) = \sum_{i,j = 1}^{n} [a_i b_j \nabla_{\frac{\partial}{\partial x_i}}(\frac{\partial}{\partial x_j})+a_i \frac{\partial b_j}{\partial x_i}\frac{\partial }{\partial x_j}]
$$
I know, that
$$
\nabla_{\frac{\partial}{\partial x_i}}(\frac{\partial}{\partial x_j}) = \sum_{k=1}^{n} \Gamma_{ij}^k \frac{\partial}{\partial x_k}
$$
on the open subspace $(U,\varphi) \in M$. Given another open subspace $(V,\phi)$, with coordinates $y_1, \ldots y_n$. Given the closed form of the covariate derivative on $V$
$$
\nabla_{\frac{\partial}{\partial y_r}}(\frac{\partial}{\partial y_s}) = \sum_{t=1}^{n} \tilde\Gamma_{rs}^t \frac{\partial}{\partial y_t}
$$
I want to show that I can express the local coefficients by the following expression:
$$
\tilde\Gamma_{rs}^t = \sum_{i,j,k=1}^{n} \frac{\partial x_i}{\partial y_r}\frac{\partial x_j}{\partial y_s}\frac{\partial y_t}{\partial x_k} \Gamma_{ij}^k + \sum_{j=1}^{n} \frac{\partial^2 x_j}{\partial y_r \partial y_s}\frac{\partial y_t}{\partial x_j}
$$
 A: We can use the fact that
$$ \frac{\partial }{\partial y_r} = \sum_i \frac{\partial x_i}{\partial y_r} \frac{\partial }{\partial x_i}.$$
Then we can use your expression for $\nabla_X(Y)$, with
$$ X = \sum_i a_i \frac{\partial }{\partial x_i}, \ \ \ \ \ a_i := \frac{\partial x_i}{\partial y_r},$$
$$ Y = \sum_j b_j \frac{\partial }{\partial x_j}, \ \ \ \ \ b_j := \frac{\partial x_j}{\partial y_s}.$$
This gives
$$ \nabla_{\frac{\partial }{\partial y_r}} \left( \frac{\partial}{\partial y_s}\right) = \sum_{i,j,k} \frac{\partial x_i}{\partial y_r}\frac{\partial x_j}{\partial y_s}\Gamma^k_{ij} \frac{\partial}{\partial x_k} + \sum_{i,j} \frac{\partial x_i}{\partial y_r} \frac{\partial}{\partial x_i} \left( \frac{\partial x_j}{\partial y_s} \right)\frac{\partial}{\partial x_j}.$$
By the chain rule,
$$ \sum_{i} \frac{\partial x_i}{\partial y_r} \frac{\partial}{\partial x_i} f = \frac{\partial f}{\partial y_r}$$
for any function $f$, so our above expression simplifies to 
$$ \nabla_{\frac{\partial }{\partial y_r}} \left( \frac{\partial}{\partial y_s}\right) = \sum_{i,j,k} \frac{\partial x_i}{\partial y_r}\frac{\partial x_j}{\partial y_s}\Gamma^k_{ij} \frac{\partial}{\partial x_k} + \sum_{j}  \frac{\partial^2 x_j}{\partial y_s \partial y_r} \frac{\partial}{\partial x_j}.$$
Finally, we use
$$ \frac{\partial }{\partial x_k} = \sum_t \frac{\partial y_t}{\partial x_k} \frac{\partial }{\partial y_t}$$
to get
$$ \nabla_{\frac{\partial }{\partial y_r}} \left( \frac{\partial}{\partial y_s}\right) = \sum_{i,j,k,t} \frac{\partial x_i}{\partial y_r}\frac{\partial x_j}{\partial y_s}\Gamma^k_{ij} \frac{\partial y_t}{\partial x_k} \frac{\partial }{\partial y_t}+ \sum_{j, t}  \frac{\partial^2 x_j}{\partial y_s \partial y_r} \frac{\partial y_t}{\partial x_j} \frac{\partial }{\partial y_t}.$$
