$$\mathbf{A} = A^{1}\mathbf{e_{1}}+A^{2}\mathbf{e_{2}}$$ $$A= \sum_{i=1}^{n}A^{i}\mathbf{e_{i}}$$
Taking the derivative wrt the tangent basis vector and dropping the summation by Einstein convention: $$\left ( \frac{\partial A^{i}}{\partial x^{j}}+\Gamma _{jk}^{i}A^{k}\right )\mathbf{e_{i}}$$
The covariant derivative of this contravector is
$$\nabla_{j}A^{i}\equiv \frac{\partial A^{i}}{\partial x^{j}}+\Gamma _{jk}^{i} A^{k}$$
Now, I would like to determine the covariant derivative of a covariant vector but ran into some problem. Namely, with the red highlighted parts in bold which does not appear in my sketch.
$$\nabla_{j}A_{i}\equiv\frac{\partial A_{i}}{\partial x^{j}}{\color{Red} -}\Gamma_{{\color{Red} ij}}^{{\color{Red} k}}A_{k}$$ which the above is a covariant derivative of a covariant vector.
Any help is appreciated.