Why don't the Christoffel symbols transform as a tensor? Let $(P, \pi, M)$ be a principal $G$ -bundle.  Given $A \in  L(G),$ we define the vector field $X^{A} \in \Gamma(T P)$ by
$$
\begin{aligned}
X_{p}^{A}: \mathcal{C}^{\infty}(P) & \stackrel{\sim}{\longrightarrow} \mathbb{R} \\
f & \mapsto[f(p\cdot\exp (t A))]^{\prime}(0),
\end{aligned}
$$
where the derivative is to be taken with respect to $t$.
Here $\Gamma$ is  the set  of section in the tangent bundle $TP$ of $(P, \pi, M)$ and $L(G)$ is the lie algebra of the group $G$.
We also define  connection one-form on the  principal bundle $(P, \pi, M)$ as a smooth Lie-algebra valued one-form, i.e. a smooth map
$$
\omega: \Gamma(T P) \stackrel{\sim}{\rightarrow} T_{e} G
$$
such that
i) $\omega\left(X^{A}\right)=A$;
ii) $\left((\triangleleft g)^{*} \omega\right)(X)=\left(\operatorname{Ad}_{g^{-1}}\right)_{*}(\omega(X))$.
Let $\sigma : U\subset M\to \pi^{-1}(U)$ be a section in $(P, \pi, M)$ the Yang–Mills field is defined as
$$
\sigma^*\omega:=A(x)=\sum_{\mu=1}^{m} \sum_{a=1}^{\operatorname{dim} G} A_{\mu}^{a}(x) E_{a}\left(d x^{\mu}\right)_{x},
$$
where $\left\{E_{1}, E_{2}, \ldots, E_{\operatorname{dim} G}\right\}$ is a basis set for $L(G)$. If $\sigma_2=g(x)\sigma$ is another section and $G$ is a matrix group then we have
$$\sigma^*\omega_2:=A_2(x)=g(x)^{-1} A_{\mu}(x) g(x)+g(x)^{-1} \partial_{\mu}g(x). \tag 1$$
Let $LM$ be a frame bundle. Any chart $(U,x)$ of a smooth manifold $M$ induces a local section $\sigma: U\to LM$ of the frame bundle of $M$ by
$$
\sigma(m):=\bigg(\frac{\partial}{\partial x^1},\ldots,\frac{\partial}{\partial x^m} \bigg)\in L_mM \tag2 
$$
we define $\Gamma:= \sigma^* \omega.$ and if   $(U',x')$ is another chart with
$U \cap U' \neq \emptyset$ and that  $$ 
\sigma_2(m):=\bigg(\frac{\partial}{\partial x'^1},\ldots,\frac{\partial}{\partial x'^m} \bigg)\in L_mM \tag3 
$$
One can show by $(1)$ that $$\Gamma_{\mu \delta}^{\prime}{ }^{\epsilon}(x)=\sum_{\alpha, \rho=1}^{m} \frac{\partial x^{\alpha}}{\partial x^{\prime \mu}} \frac{\partial x^{\rho}}{\partial x^{\prime \delta}} \frac{\partial x^{\prime \epsilon}}{\partial x^{\chi}} \Gamma_{\alpha \rho}^{\chi}(x)+\sum_{\lambda=1}^{m} \frac{\partial x^{\prime \epsilon}}{\partial x^{\lambda}} \frac{\partial^{2} x^{\lambda}}{\partial x^{\prime \mu} \partial x^{\prime \delta}}.\tag 4$$
As we see expression $(4)$ does not come from  a coordinate transformation but from a change of section $\sigma \rightarrow \sigma_2$.
Now for charts $(U,x)$ and $(U,x')$ if one makes a coordinate transformation we would have $$\sigma(x) \rightarrow \sigma'(x')=\bigg(\frac{\partial x'^{\nu}}{\partial x^{1}}\frac{\partial}{\partial x'^{\nu}},\ldots,\frac{\partial x'^{\mu}}{\partial x^{m}}\frac{\partial}{\partial x'^{\mu}} \bigg)= \bigg(\frac{\partial}{\partial x^1},\ldots,\frac{\partial}{\partial x^m} \bigg)=\sigma(x) \tag 5 $$ and so
$\Gamma(x)=\Gamma'(x')$ and since only the indices $\mu$ depends  on the coordinates  we should have
$$\Gamma_{\mu \delta}^{\prime}{ }^{\epsilon}=\frac{\partial x'^{\nu}}{\partial x^{\mu}}\Gamma_{\nu \delta}{ }^{\epsilon}.$$
Am I wrong?
 A: Let $M$ be a smooth manifold and $(P,\pi,M,G)$ a principal $G$-bundle over $M$. Let $\omega$ be a connection one-form and let $\sigma : U\subset M\to \pi^{-1}(U)$ be a local section which locally trivializes the bundle by means of $h:U\times G\to \pi^{-1}(U)$:
$$h(x,g)=\sigma(x)\cdot g.\tag{1}$$
With $h$ we can pullback $\omega$ to understand how it is represented on the locally trivial neighborhood. What you must do is evaluate $h^\ast \omega$. This is done in Theorem 6.1 of "Modern Differential Geometry for Physicists" by C. J. Isham, and the result is, after identifying $T_{(x,g)}(U\times G)\simeq T_x U\oplus T_gG$: $$(h^\ast \omega)_{(x,g)}(\alpha,\beta)={\rm Ad}_{g^{-1}}((\sigma^\ast\omega)_x(\alpha))+\Xi_g(\beta)\tag{2},$$
where $g\mapsto {\rm Ad}_{g}$ is the adjoint representation of $G$ on the Lie algebra $\mathfrak{g}$ and $\Xi$ is the Maurer-Cartan form. Observe that a connection one-form is locally specified by a $\mathfrak{g}$-valued one form in $U$, but there is also the Maurer-Cartan part, this will play a role in a second.
Now, $\sigma^\ast\omega$ is a $\mathfrak{g}$-valued one-form, and it transforms as such under changes of coordinates on $U$, but that is not the point. The point is that if you pick a distinct section $\widetilde{\sigma}:V\subset M\to \pi^{-1}(V)$ with $U\cap V\neq\emptyset$ you get a distinct local trivialization $\widetilde{h}:V\times G\to \pi^{-1}(V)$ and from (2) it gives you a diferent $\mathfrak{g}$-valued one-form $\widetilde{\sigma}^\ast \omega$.
If you restrict to $U\cap V$ you have two $\mathfrak{g}$-valued one forms on this open subset: $\sigma^\ast\omega$ and $\widetilde{\sigma}^\ast\omega$. Then you would like to relate these two and you find
out that the Maurer-Cartan form, as could be expected, plays a role. Indeed, defining $\Omega:U\cap V\to G$ by $\widetilde{\sigma}(x)=\sigma(x)\Omega(x)$, this is shown in Theorem 6.2 of Isham's book:
$$(\widetilde{\sigma}^\ast\omega)_\mu(x)={\rm{Ad}}_{\Omega(x)^{-1}}((\sigma^\ast\omega)_\mu(x))+(\Omega^\ast\Xi)_\mu(x).\tag{3}$$
As it happens for the transformation law of the local representative of a connection you have an inhomogeneous part, which here we see that corresponds to the Maurer-Cartan form. This is the origin of the inhomogeneous term in the transformation law for the Christoffel symbols.
The point IMHO is that in this analysis we are not talking about the transformation of the components of a single one-form but rather we are talking about how two distinct one-forms are related to one another given that they are distinct representatives of the same fundamental object on the principal bundle.
Finally for the specific version of this formalism with ${\rm GL}(m,\mathbb{R})$ see section 6.1.4 of Isham's book.
A: 
I am not seeing how since the only part of ${\Gamma_j^i}_\mu$ which depends on the coordinates is the index $\mu$ which transform as one form.

That's not accurate.  You're right that the $\mu$ index refers to the manifold coordinates and the $i$ and $j$ indices refer to the tangent space.  However, the tangent space basis vectors also depend on the manifold coordinates.  In other words, $\boldsymbol \Gamma \equiv (\Gamma_1,\ldots,\Gamma_{\mathrm{dim}(M)})$ is a $T_e\mathrm{GL}(m,\mathbb R)$-valued one-form (i.e. a collection of matrices), but $\big(\Gamma_{\mu}\big)^i_{\ j}$ are the components of the matrix $\Gamma_\mu$ in the local tangent space basis.

The connection defines a parallel transport $T$ for tangent vectors.  If we transport the basis vector $\hat e_j$ an infinitesimal distance $\epsilon$ along the $x^\mu$ direction, we obtain
$$T(x^\mu, \epsilon)\hat e_j = (\mathbb I + \epsilon \Gamma_\mu)\hat e_j$$
If $\Gamma_\mu$ vanishes, then that implies that the $j^{th}$ tangent vector at the point $(x^1,\ldots,x^N)$ is parallel-transported to itself at the point $(x^1,\ldots,x^\mu+\epsilon,\ldots,x^N)$ along the specified curve.  Loosely, we would say that the basis vectors do not depend on the coordinate $x^\mu$.
However, even if this is the case in one choice of basis, there is no reason whatsoever that this should be true in any other basis.  If we choose a basis which varies with position (such as the coordinate-induced basis in curvilinear coordinates), then $\mathbf \Gamma \neq 0$.

In summary, the non-tensorial transformation properties of $\Gamma_\mu)^i_{\ j}$ arise because $\Gamma_\mu$ measures the rate of change of the basis vectors along the direction $x^\mu$.  If you're using the coordinate-induced basis for the tangent bundle, then a change of coordinates induces a corresponding change of basis.  The spatial rates of change of those basis vectors are clearly non-tensorial - one can see this by noting that their vanishing in one basis does not imply their vanishing in any other basis.  As a result, the components $(\Gamma_\mu)^i_{\ j}$ do not transform as a $(1,2)$-tensor.


As we see expression (4) does not come from a coordinate transformation but from a change of section $\sigma\rightarrow\sigma_2$.

Yes, but the section is induced by the coordinate chart.

Now for charts $(U,x)$ and $(U,x')$ if make a coordinate transformation we would have $\sigma(x)=\sigma'(x')$ [...]

This isn't true.  It should be clear that if $x$ and $x'$ are different charts, then $\left\{\frac{\partial}{\partial x^\mu}\right\} \neq \left\{\frac{\partial}{\partial x'^\mu}\right\}$.  They may have the same span, but that's different from being the same frame.
If you're using the coordinate-induced basis for your tangent spaces, then a change of coordinates is accompanied by a change in frame which ultimately results in the non-tensorial transformation for the connection coefficients.
It appears that you are trying to change coordinates without changing the frame, e.g. transforming from cartesian to polar coordinates but continuing to use the cartesian basis vectors as the basis for the tangent bundle.  In principle this is possible, and the connection coefficients would transform as you say.  But when we say that the connection coefficients do not transform as a tensor, we refer to coordinate changes and the corresponding induced change in coordinate frame.

$$\sigma(x) \rightarrow \sigma'(x')=\bigg(\frac{\partial x'^{\nu}}{\partial x^{1}}\frac{\partial}{\partial x'^{\nu}},\ldots,\frac{\partial x'^{\mu}}{\partial x^{m}}\frac{\partial}{\partial x'^{\mu}} \bigg)= \bigg(\frac{\partial}{\partial x^1},\ldots,\frac{\partial}{\partial x^m} \bigg)=\sigma(x)$$

This isn't true.  $\sigma'(x')$ - the frame induced by the chart $x'$ - is simply $\left(\frac{\partial}{\partial x'^1} ,\ldots,\frac{\partial}{\partial x'^m}\right)$ by the definition you yourself provide in (3).
A: The Christoffel symbols are not the components of a tensor. Their transformation laws involve the second derivatives of the parameter transformation :
\begin{align}
\bar{\Gamma}^k_{ij}&\boldsymbol{=} \left[\Gamma^\gamma_{\alpha\beta}\dfrac{\partial u^\alpha}{\partial\bar{u}^i}\dfrac{\partial u^\beta}{\partial\bar{u}^j}\boldsymbol{+}\dfrac{\partial^2 u^\gamma}{\partial\bar{u}^i\partial\bar{u}^j}\right]\dfrac{\partial \bar{u}^k}{\partial u^\gamma}\boldsymbol{\ne} \Gamma^\gamma_{\alpha\beta}\dfrac{\partial u^\alpha}{\partial\bar{u}^i}\dfrac{\partial u^\beta}{\partial\bar{u}^j}\dfrac{\partial \bar{u}^k}{\partial u^\gamma}
\tag{01}\label{01}\\
\bar{\Gamma}_{ijk}&\boldsymbol{=} \left[\Gamma_{\alpha\beta\gamma}\dfrac{\partial u^\alpha}{\partial\bar{u}^i}\dfrac{\partial u^\beta}{\partial\bar{u}^j}\boldsymbol{+}g_{\alpha\gamma}\dfrac{\partial^2 u^\alpha}{\partial\bar{u}^i\partial\bar{u}^j}\right]\dfrac{\partial u^\gamma}{\partial \bar{u}^k}\boldsymbol{\ne} \Gamma_{\alpha\beta\gamma}\dfrac{\partial u^\alpha}{\partial\bar{u}^i}\dfrac{\partial u^\beta}{\partial\bar{u}^j}\dfrac{\partial u^\gamma}{\partial \bar{u}^k}
\tag{02}\label{02}
\end{align}
