Expressions for components of a covariant derivative of a vector and a covector I'm trying study by myself Riemannian Geometry and Tensor Calculus in order to be able to understand these notes about Mean Curvature Flow. I'm on page $4$ of the notes and I'm stuck in understand why the expressions for covariant derivative of a vector $v$ and a covector $w$ are
$$\nabla_k v^i = \frac{\partial v^i}{x_k} + \Gamma^i_{jk} v^j$$
and
$$\nabla_k w_j = \frac{\partial w_j}{x_k} - \Gamma^i_{jk} w_i,$$
respectively, where $v^i$ and $w_j$ denote the components of the vector $v$ and the covector $w$ respectively. I know that is adopted in Riemannian Geometry the Einstein Summation Convention, but I will write the sum and denote by $\partial_i := \frac{\partial}{\partial x_i}$ in my attempt to follow:
$\textbf{My attempt:}$
Define $v' := \nabla_k (v^i \partial_i)$.
$v' := \nabla_k (v^i \partial_i) = \left( \nabla_k v^i \right) \partial_i + v^i \left( \nabla_k \partial_i \right) = \frac{\partial v^i}{\partial x_k} \partial_i + v^i \sum_{j=1}^{j=n} \Gamma^i_{jk} \partial_j = \left( \frac{\partial v^i}{\partial x_k}  + v^i \Gamma^i_{ik} \right) \partial_i +  \sum_{j=1, j \neq i}^{j=n} \Gamma^i_{jk} \partial_j$, 
then the $i^{th}$-coordinate of $v'$ written in coordinate basis is $(v')^i := \frac{\partial v^i}{\partial x_k}  + v^i \Gamma^i_{ik}$.
Define $w':= \nabla_k w_j$ and observe that
$dx^j \left( \partial_j \right) = 1 \implies \nabla_k \left( dx^j \left( \partial_j \right) \right) = \nabla_k 1 \implies \nabla_k \left( dx^j \right) \left( \partial_j \right) + dx^j \left( \nabla_k \partial_j \right) = 0 \implies \nabla_k \left( dx^j \right) \left( \partial_j \right) + dx^j \left( \sum_{i=1}^{i=n} \Gamma^i_{jk} \partial_i \right) = 0 \implies \nabla_k \left( dx^j \right) \left( \partial_j \right) + \Gamma^j_{jk} = 0 \implies \nabla_k \left( dx^j \right) \left( \partial_j \right) = - \Gamma^j_{jk}$
Therefore,
$\nabla_k w_j = \nabla_k (w_ j \ 1) = \nabla_k \left( w_j dx^j \left( \partial_j \right) \right) = \nabla_k \left( w_j \right) dx^j  \left( \partial_j \right) + w_j \nabla_k \left( dx^j \left( \partial_j \right) \right) = \frac{\partial w_j}{\partial x_k} - \Gamma^j_{jk} w_j$
In the two cases, I didn't obtain the expressions that I would like, where am I going wrong? Anyone can help me to obtain the expressions for covariant derivative of a vector $v$ and a covector $w$?
$\textbf{P.S.:}$ I used the product rule to derive tensors given by Barret Oneill in his book: Semi-Riemannian Geometry with applications to relativity.
Thanks in advance!
 A: Let's do a simple example to see the full calculation, without using the sum. Your first problem is that you should have $ \nabla_k(\partial_j) = \Gamma^i_{kj} \partial_i $, or $\sum_{i} \Gamma^i_{kj} \partial_i $ if you prefer, so that the indices are only summed when it appears once as an upper and once lower. (This is the definition of the Christoffel symbols.)
Then, putting every summation in, you find 
$$ \nabla_k \left( \sum_i v^i \partial_i \right) = \sum_i \left( ( \partial_k v^i )\partial_i + v^i \nabla_k(\partial_i) \right) = \sum_i  (\partial_k v^i)\partial_i + \sum_i\sum_{m} v^i \Gamma^m_{ki} \partial_m . $$
Here I think is your second problem: there are two sums over basis vectors with indices, so you either need to change the dummy indices in the sums so that you extract the same vector from both, or dot with something in the dual basis. The latter gives
$$ dx^j \left( \nabla_k \left( \sum_i v^i \partial_i \right) \right) = \sum_i  (\partial_k v^i) dx^j(\partial_i) + \sum_i\sum_{m} v^i \Gamma^m_{ki} dx^j(\partial_m) \\ 
= \sum_i  (\partial_k v^i) \delta^j_i + \sum_i\sum_{m} v^i \Gamma^m_{ki} \delta^j_m \\
=   \partial_k v^j + \sum_{m} v^i \Gamma^j_{ki}, $$
so this is the $j$th component.
Regarding the second part, in fact $dx^j(\partial_j)=n$, the dimension (there is an implied sum over $j$ here). I think you want to start from $dx^i(\partial_j)=\delta^i_j$, which is also constant, so the rest of the proof will work:
$$ 0 = \nabla_{k} \delta^i_j = \nabla_k(dx^i(\partial_j)) = (\nabla_k dx^i)(\partial_j) + dx^i(\nabla_k \partial_j) = (\nabla_k dx^i)(\partial_j) + dx^i\left(\sum_m\Gamma^{m}_{kj} \partial_m\right) = (\nabla_k dx^i)(\partial_j) + \Gamma^i_{kj}, $$
and so on.
