Two basic questions about Christoffel symbols I am trying to understand (rather than memorise) the derivation of the Christoffel symbols from the vanishing covariant derivative of the metric, the very first step is
\begin{equation}
\label{eq:first}
\nabla_\sigma g_{\mu\nu}=\partial_\sigma g_{\mu\nu}-\Gamma^\lambda_{\sigma\mu}g_{\lambda\nu}-\Gamma^\lambda_{\sigma\nu}g_{\mu\lambda}=0.
\end{equation}
I'm wondering what is a good way to remember (or better yet work out) how to index the $\Gamma$ coefficients? How to know which indexes are subscripted on the $\Gamma$ and which index on the metric is involved in the summation with the upstairs index on the $\Gamma$? Is there a consistent way to work these out? Because at the moment I take at face value what is written in notes which is not very satisfying and feels like keeping the training wheels on.

The second issue I have is in the Leibniz when taking the covariant derivative of the metric. The metric is a rank-2 tensor and so can be written in the form
\begin{equation}
g_{\mu\nu}\tilde e^\mu \otimes \tilde e^\nu.
\end{equation}
When we take the covariant derivative we apply the product rule first (I assume) and then use Leibniz rule to take the derivatives of the dual basis vectors, i.e:
\begin{equation}
\nabla_\sigma g_{\mu\nu}\tilde e^\mu \otimes \tilde e^\nu=(\partial g_{\mu\nu})\tilde e^\mu \otimes \tilde e^\nu+g_{\mu\nu}\bigl((\nabla_\sigma\tilde e^\mu)\otimes\tilde e^\nu\bigr)+g_{\mu\nu}\bigl((\nabla_\sigma\tilde e^\nu)\otimes\tilde e^\mu\bigr)\end{equation}
My question is, is this the correct method? It feels wrong to be taking the tensor product of the connection coefficient this way given that it isn't a true tensor. And if this is correct, where does the tensor product fit into Eq.1?
 A: 1. I believe you have read all the corresponding info about Christoffel symbols on wikipedia and all you need is a good mnemonic. We need to deal with two things: indices and sign. The index part is actually simple. If you have a covector:
$$
\nabla_iu_j=\partial_iu_j-\Gamma^{\square}_{\square\square}u_{\square}
$$
then it's obvious that $\Gamma$ should contract with $u$, in other words the dummy index should go to $u$ and to top of $\Gamma$. The two lower free indices should go to bottom of $\Gamma$:
$$
\nabla_iu_j=\partial_iu_j-\Gamma^{k}_{ij}u_{k}
$$
For vector:
$$
\nabla_iv^j=\partial_iv^j +\Gamma^{\square}_{\square\square}v^{\square}
$$
it's again obvious that $v$ should have a dummy index and it doesn't matter to which of the lower indices of $\Gamma$ it goes, since $\Gamma$ is symmetric on lower indices (however, traditionally dummy index comes first). Free indices should go to corresponding places: one up, one down
$$
\nabla_iv^j=\partial_iv^j +\Gamma^j_{ki}v^k.
$$
Generally, if you have an arbitrary tensor:
$$
\nabla_iT_{\color{red}{l}\color{green}m\ldots}^{\color{magenta}p\color{orange}q\ldots}=\partial_iT_{lm\ldots}^{pq\ldots}
\color{red}{-\Gamma^k_{il}T}_{\color{red}km\ldots}^{pq\ldots}
\color{green}{-\Gamma^k_{im}T}_{l\color{green}k\ldots}^{pq\ldots} \ldots
\color{magenta}{+ \Gamma^p_{ki}T}_{lm\ldots}^{\color{magenta}kq\ldots}
\color{orange}{+ \Gamma^q_{ki}T}_{lm\ldots}^{p\color{orange}k\ldots} \ldots
$$
for each of the index you add a term as if this thing was just a vector or a covector.
Finally, the sign part is easy to remember since “vector-covector” and “plus-minus” are both naturally ordered.
2. That's a right approach. When you substitute
$$
\frac{\partial\mathbf e^\mu}{\partial x^\sigma} = -\Gamma^\mu_{\sigma\nu}\mathbf e^\nu,
$$
you will obtain the equation above.
