This case is not a proper one to understand the parallel transport and the role of the Christoffel Symbols. This is due to the fact that the latter are all identically zero.
For the parallel transport of a vector $\:V^\alpha\:$ along a curve with parametric equation
\begin{equation}
\boldsymbol{p}\left(\lambda\right) \boldsymbol{=}\left[p^1 \left(\lambda\right), p^2 \left(\lambda\right)\right]\qquad \left(p^1\boldsymbol{\equiv}p,\:p^2\boldsymbol{\equiv}q\right)
\tag{01}\label{01}
\end{equation}
we use the equation
\begin{equation}
\frac{\mathrm dV^\alpha}{\mathrm d\lambda} \boldsymbol{=} \boldsymbol{-}\Gamma_{\beta\nu}^{\alpha}V^\nu\frac{\mathrm dp^{\,\beta}}{\mathrm d\lambda} \qquad \left(\alpha,\beta,\nu \boldsymbol{=}1,2\right)
\tag{02}\label{02}
\end{equation}
or equivalently
\begin{equation}
\mathrm dV^\alpha \boldsymbol{=} \boldsymbol{-}\Gamma_{\beta\nu}^{\alpha}V^\nu \mathrm dp^{\,\beta} \qquad \left(\alpha,\beta,\nu \boldsymbol{=}1,2\right)
\tag{03}\label{03}
\end{equation}
The Christoffer symbols are expressed through the components of the metric tensor $\mathbf{g}$
\begin{equation}
\Gamma_{\beta\nu}^{\alpha}\boldsymbol{=} \frac{1}{2}g^{\alpha k} \left(\frac{\partial g_{k\nu}}{\partial p^{\,\beta }}\boldsymbol{+}\frac{\partial g_{\beta k}}{\partial p^{\,\nu}}\boldsymbol{-}\frac{\partial g_{\beta \nu}}{\partial p^{\,k}}\right)
\tag{04}\label{04}
\end{equation}
In all equations we make use of the Einstein summation convention.
The metric tensor is determined from the expression of the infinitesimal displacement
\begin{equation}
\mathrm ds^2 \boldsymbol{=}g_{\rho\sigma}\mathrm dp^{\,\rho}\mathrm dp^{\,\sigma}
\tag{05}\label{05}
\end{equation}
In our case
\begin{equation}
\mathrm ds^2 \boldsymbol{=}\mathrm d\mathbf r\boldsymbol{\cdot}\mathrm d\mathbf r\boldsymbol{=}\left(\mathbf e_1\mathrm dp\boldsymbol{+}\mathbf e_2\mathrm dq\right)\boldsymbol{\cdot}\left(\mathbf e_1\mathrm dp\boldsymbol{+}\mathbf e_2\mathrm dq\right) \boldsymbol{=}g_{11}\mathrm dp^{2}\boldsymbol{+}2g_{12}\mathrm dp\mathrm dq\boldsymbol{+}g_{22}\mathrm dq^{2}
\tag{06}\label{06}
\end{equation}
explicitly
\begin{equation}
\mathrm ds^2 \boldsymbol{=}\overbrace{\underbrace{\left(\mathbf e_1\boldsymbol{\cdot}\mathbf e_1\right)}_{g_{11}}}^{\tfrac{1}{2}}\mathrm dp^{2}\boldsymbol{+}2\overbrace{\underbrace{\left(\mathbf e_1\boldsymbol{\cdot}\mathbf e_2\right)}_{g_{12}}}^{\boldsymbol{-}2}\mathrm dp\mathrm dq\boldsymbol{+}\overbrace{\underbrace{\left(\mathbf e_2\boldsymbol{\cdot}\mathbf e_2\right)}_{g_{22}}}^{\tfrac{17}{2}}\mathrm dq^{2}
\tag{07}\label{07}
\end{equation}
where as the OP gives
\begin{equation}
\mathbf e_1 \boldsymbol{=}\frac12\left(\boldsymbol{-}\mathbf e_x\boldsymbol{+}\mathbf e_y\right)\,,\qquad \mathbf e_2 \boldsymbol{=}\frac12\left(5\mathbf e_x\boldsymbol{-}3\mathbf e_y\right)
\tag{08}\label{08}
\end{equation}
So the metric tensor is
\begin{equation}
\mathbf g\boldsymbol{=}g_{\rho\sigma}\boldsymbol{=}
\begin{bmatrix}
g_{11} & g_{12} \vphantom{\dfrac{a}{b}}\\
g_{21} & g_{22} \vphantom{\dfrac{a}{b}}
\end{bmatrix}
\boldsymbol{=}
\begin{bmatrix}
\hphantom{\boldsymbol{-}}\tfrac{1}{2} & \boldsymbol{-}2 \vphantom{\dfrac{a}{b}}\\
\boldsymbol{-}2 & \hphantom{\boldsymbol{-}}\tfrac{17}{2} \vphantom{\dfrac{a}{b}}
\end{bmatrix}
\tag{09}\label{09}
\end{equation}
The metric tensor is independent of the $''$curvilinear$''$ coordinates $p,q$ so all Christoffel Symbols are identically zero
\begin{equation}
\frac{\partial g_{\rho\sigma}}{\partial p^{\tau }}\boldsymbol{=}0\quad \boldsymbol{\implies} \quad \Gamma_{\beta\nu}^{\alpha}\boldsymbol{=} \frac{1}{2}g^{\alpha k} \left(\frac{\partial g_{k\nu}}{\partial p^{\,\beta }}\boldsymbol{+}\frac{\partial g_{\beta k}}{\partial p^{\,\nu}}\boldsymbol{-}\frac{\partial g_{\beta \nu}}{\partial p^{\,k}}\right)\boldsymbol{=}0
\tag{10}\label{10}
\end{equation}
and from equation \eqref{03}
\begin{equation}
\mathrm dV^\alpha \boldsymbol{=} 0 \qquad \left(\alpha \boldsymbol{=}1,2\right)
\tag{11}\label{11}
\end{equation}
