I'm not exactly sure which "derivation" you're referring to, but I've always seen Christoffel symbols introduced in the following context:
Ultimately the motivation is to define the notion of parallel transport, which means when we take a derivative of a vector field, we want our derivative to still be tangent to our manifold. So let's suppose for instance we have a manifold $M$, and we have vector fields $X, Y \in \mathcal{X}(M)$ defined locally in $M$. We want an affine connection $\nabla:\mathcal{X}(M) \times \mathcal{X}(M) \to \mathcal{X}(M)$ to act as a derivative that remains tangent to the manifold. Let $\frac{\partial}{\partial e^1}, \ldots, \frac{\partial }{\partial e^n}$ be a local basis for the tangent spaces on $M$ (defined in the same neighborhood as our vector fields), such that we can express our vector fields locally by functions $x^i, y^j: M \to \mathbb{R}$:
$$
X \;\; =\;\; x^i \frac{\partial}{\partial e^i} \;\;\; \;\;\; Y \;\; =\;\; y^j \frac{\partial}{\partial e^j}.
$$
Then our covariant derivative is given locally by
$$
\nabla_XY \;\; =\;\; \nabla_{x^i\partial_i} (y^j \partial_j) \;\; =\;\; x^i \nabla_{\partial_i} (y^j\partial_j) \;\; =\;\; x^i \frac{\partial y^j}{\partial e^i} \frac{\partial }{\partial e^j} + x^iy^j \nabla_{\partial_i}\partial_j.
$$
The first term is clearly in the tangent space, but we want to define $\nabla_{\partial_i}\partial_j$ to lie in the tangent space. We therefore define correction functions $\Gamma_{ij}^k: M\to \mathbb{R}$ known as the Christoffel symbols that satisfy
$$
\nabla_{\partial_i} \partial_j \;\; \equiv \;\; \Gamma_{ij}^k \frac{\partial}{\partial e^k}.
$$
Now by manipulating the indices around a little bit we have that
$$
\nabla_XY \;\; =\;\; \left (x^i \frac{\partial y^k}{\partial e^i} + x^i y^j \Gamma_{ij}^k \right ) \frac{\partial}{\partial e^k}.
$$
The above analysis is valid for any affine connection on a manifold, but on a Riemannian manifold we require the connection to be symmetric and compatible with the metric. The compatibility condition is that
$$
X\langle Y,Z\rangle \;\; =\;\; \langle \nabla_XY, Z\rangle + \langle Y, \nabla_XZ\rangle
$$
while the symmetry condition is that
$$
\nabla_XY - \nabla_YX \;\; =\;\; [X,Y]
$$
which for the basis vectors reduces to $\nabla_{\partial_i}\partial_j - \nabla_{\partial_j}\partial_i = 0$. From these equations we can in general prove that
$$
\langle Z, \nabla_YX\rangle \;\; =\;\; \frac{1}{2} \left ( X\langle Y,Z\rangle + Y\langle Z, X\rangle - Z \langle X, Y\rangle - \langle [X,Z], Y\rangle - \langle [Y,Z], X\rangle - \langle [X,Y], Z \rangle \right ).
$$
If we reduce this down to the basis vectors we obtain
$$
\langle \partial_s, \nabla_{\partial_i}\partial_j\rangle \;\; =\;\; \Gamma_{ij}^k g_{sk} \;\; =\;\; \frac{1}{2} \left ( \partial_j g_{si} + \partial_i g_{sj} - \partial_s g_{ij} \right ).
$$
Because we have that $g^{sm}$ is the inverse of $g_{sk}$ we have that $g_{sk} g^{sm} = \delta_k^m$ and thus we obtain
$$
\Gamma_{ij}^m \;\; =\;\; \frac{1}{2} g^{ms} \left ( \partial_j g_{si} + \partial_i g_{sj} - \partial_s g_{ij} \right ).
$$
This is the intuitive explanation that I like most regarding Christoffel symbols: they are correction functions to make sure that an affine connection keeps the derivative of a vector field within the tangent space of the manifold. On a Riemannian manifold they have the given form above. To get the other equations that you have listed in your question, simply permute the indices to get different copies of the fundamental equation for $\Gamma_{ij}^s$, and then subtract/add them together to get the forms you're interested in.
Best Answer
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.