A connection $\nabla$ on $TM$ can be defined abstractly as a map $\nabla: \mathfrak X(M) \times \mathfrak X(M) \to \mathfrak X(M), (X,Y) \mapsto \nabla_X Y$, where $\mathfrak X(M)$ is the set of vector fields on $X$, such that $\nabla$ is $C^\infty(M)$ linear in the $X$ variable, $\mathbb R$ linear in $Y$, and satisfies the Leibniz rule
$$
\nabla_X fY = df(X) Y + f\nabla_X Y
$$
for smooth functions $f$. If $\partial_{x_i}$ is a local coordinate chart for $TM$ then it is easy to see that $\nabla$ is completely determined by the vector fields $\nabla_{\partial_{x_i}} \partial_{x_j}$. Then $\Gamma_{ij}^k$ is by definition the coordinate functions of $\nabla_{\partial_{x_i}}\partial_{x_j}$. Note that these coefficients do not piece together to form a tensor since a connection is not tensorial (it is not $C^\infty$ linear in $Y$).
Now, by definition, the Levi-Civita connection is the unique connection on $TM$ that has no torsion and is compatible with the metric. From these properties it can be shown that the Christoffel symbols of this connection are given by the first equation you gave.
This is very straightforward, just substitute the transformation rules and collect the terms.
Here are some details.
The inverse metric transforms, as we know, by the rule:
$$
g^{\mu \lambda} = \frac{\partial{\bar{x}}^\mu}{\partial{x}^\alpha} \frac{\partial{\bar{x}}^\lambda}{\partial{x}^\delta} g^{\alpha \delta}
$$
The partial derivatives need some calculations that can be presented as
$$
\begin{align*}
g_{\lambda \kappa , \nu} & = \frac{\partial}{\partial{\bar{x}^\nu}} \Big( \frac{\partial{x^\delta}}{\partial{\bar{x}^\lambda}} \frac{\partial{x^\gamma}}{\partial{\bar{x}^\kappa}} g_{\delta \gamma} \Big) \\
&= \frac{\partial{x^\delta}}{\partial{\bar{x}^\lambda}} \frac{\partial{x^\gamma}}{\partial{\bar{x}^\kappa}} \frac{\partial{x^\beta}}{\partial{\bar{x}^\nu}} g_{\delta \gamma , \beta} + g_{\delta \gamma} \frac{\partial}{\partial{\bar{x}^\nu}} \Big( \frac{\partial{x^\delta}}{\partial{\bar{x}^\lambda}} \frac{\partial{x^\gamma}}{\partial{\bar{x}^\kappa}} \Big)
\end{align*}
$$
Similarly,
$$
g_{\nu \lambda , \kappa} = \frac{\partial{x^\beta}}{\partial{\bar{x}^\nu}} \frac{\partial{x^\delta}}{\partial{\bar{x}^\lambda}} \frac{\partial{x^\gamma}}{\partial{\bar{x}^\kappa}} g_{\beta \delta , \gamma} + g_{\beta \delta} \frac{\partial}{\partial{\bar{x}^\kappa}} \Big( \frac{\partial{x^\beta}}{\partial{\bar{x}^\nu}} \frac{\partial{x^\delta}}{\partial{\bar{x}^\lambda}} \Big)
$$
and
$$
g_{\nu \kappa , \lambda} = \frac{\partial{x^\beta}}{\partial{\bar{x}^\nu}} \frac{\partial{x^\gamma}}{\partial{\bar{x}^\kappa}} \frac{\partial{x^\delta}}{\partial{\bar{x}^\lambda}} g_{\beta \gamma , \delta} + g_{\beta \gamma} \frac{\partial}{\partial{\bar{x}^\lambda}} \Big( \frac{\partial{x^\beta}}{\partial{\bar{x}^\nu}} \frac{\partial{x^\gamma}}{\partial{\bar{x}^\kappa}} \Big)
$$
Substituting these identities into your "definition"
$$
\Gamma^\mu _{\nu\kappa} = \frac{1}{2}g^{\mu\lambda}\left(g_{\lambda\kappa,\nu}+g_{\nu\lambda,\kappa}-g_{\nu\kappa,\lambda} \right)
$$
and taking into account that
$$
\Gamma^\alpha _{\beta \gamma} = \frac{1}{2}g^{\alpha \delta}\left(g_{\delta \gamma , \beta}+g_{\beta \delta , \gamma} - g_{\beta \gamma , \delta} \right)
$$
it is not difficult now to show the required transformation rule for the Christoffel symbols.
Best Answer
$\def\+#1{\frac{\partial}{\partial u^{#1}}}$We have, as the metric has vanishing covariant derivative \[ 0 = \nabla_{\+j} g_{ik} = \+j g_{ik} - \sum_m (g_{im}\Gamma^m_{kj} + g_{mk}\Gamma^m_{ij}) \] or \[ \+jg_{ik} = \sum_m (g_{im}\Gamma^m_{kj} + g_{mk}\Gamma^m_{ij})\] That is \begin{align*} \frac 12 \sum_l g^{kl}\left( \+j g_{il} + \+ig_{jl} - \+l g_{ij}\right) &= \frac 12 \sum_{m,l} g^{kl} \left(g_{im}\Gamma^m_{lj} + g_{ml}\Gamma^m_{ij} + g_{jm}\Gamma^m_{li} + g_{ml}\Gamma^m_{ji} - g_{im}\Gamma^m_{jl} - g_{mj}\Gamma^m_{il}\right)\\ &= \frac 12 \sum_{m,l} \left(g^{kl}g_{im}\Gamma^m_{jl} - g^{kl}g_{im}\Gamma^m_{lj}\right) + \frac 12(\Gamma_{ij}^k + \Gamma^k_{ji})\\ &= \Gamma^k_{ij} \quad\text{by symmetry} \end{align*}