And the Connection fellow enters the stage:
So, an (affine) connection is the map:
$$\begin{array}{rl} \nabla: \chi(\mathcal{M}) \times \chi(\mathcal{M}) &\to \chi(\mathcal{M}) \\ (V,W) &\mapsto \nabla_{V}W \end{array} $$
with the axioms:
$1. \nabla_{V}(W+Z) = \nabla_{V}(W) + \nabla_{V}(Z)$
$2.\nabla_{V+Z}(W) = \nabla_{V}(W) + \nabla_{Z}(W)$
$3.\nabla_{fV}(W) = f\nabla_{V}(W)$
$4.\nabla_{V}(fW) = V[f]W+f \nabla_{V}(W)$
Now, given a connection we can take the tangent vector field $\dot{\gamma} = \dot{t}^{i}\frac{\partial}{\partial x^{i}}$ to a given curve $\gamma(\lambda)$ of the manifold, and ask ourselfs about the change of the vector field $V$ along the curve $\gamma(\lambda)$: $\nabla_{\dot{\gamma}} V$ .
Now, in a given chart, we whole thing is written as
\begin{equation} \nabla _{\dot{\gamma}} V = \Bigg[\dot{t}^{i}\Big(\partial_{i}v^{k} + \Gamma^{k}\hspace{0.1mm}_{ij} v^{j}\Big)\Bigg]\partial_{k} \tag{1} \end{equation}
But in general case we have:
\begin{equation} \nabla _{X} V = \Bigg[X^{i}\Big(\partial_{i}v^{k} + \Gamma^{k}\hspace{0.1mm}_{ij} v^{j}\Big)\Bigg]\partial_{k} \tag{2} \end{equation}
Furthermore, since we can define (at this point is actually just a mere change of symbology):
$$ \nabla_{i}v^{k}= \partial_{i}v^{k} + \Gamma^{k}\hspace{0.1mm}_{ij} v^{j} \tag{3}$$
we note that we have covariant and a contravariant indexes in a new object:
$$T^{k}_{i} := \nabla_{i}v^{k} = \partial_{i}v^{k} + \Gamma^{k}\hspace{0.1mm}_{ij} v^{j} \tag{4}$$
After we change the coordinates we realize that:
$$ \nabla_{a'}v^{b'} = \frac{\partial x^{b'}}{\partial x^{k}}\frac{\partial x^{i}}{\partial x^{a'}}\nabla_{i}v^{k} \tag{5}$$
Therefore we can define:
$$\nabla V := (\nabla_{i}v^{k}) dx^{i} \otimes \frac{\partial}{\partial x^{k}} \tag{6}$$
So, we have then at least three objects: $\nabla _{\dot{\gamma}} V$, $\nabla _{X} V$, $\nabla V$. And, another one that we use to define geodesics: $\nabla _{\dot{\gamma}} \dot{\gamma}$:
\begin{equation} \nabla_{\dot{\gamma}}\dot{\gamma} = \Bigg( \frac{d^{2}x^{k}}{d\lambda^{2}} + \Gamma^{k}\hspace{0.1mm}_{ij}\frac{dx^{i}}{d\lambda} \frac{dx^{j}}{d\lambda} \Bigg)\partial_{k} \tag{7} \end{equation}
My question is: is it legal to construct the tensor $(6)$ the way that I did?(*)
(*)I'm asking this because, in General Relativity text books, we call a covariant derivative precisely the components: $\nabla_{i}v^{k} = \partial_{i}v^{k} + \Gamma^{k}\hspace{0.1mm}_{ij} v^{j}$ and not the $X^{i}(\nabla_{i}v^{k}) = X^{i}(\partial_{i}v^{k} + \Gamma^{k}\hspace{0.1mm}_{ij} v^{j})$ (which are precisely the components of $\nabla_{X}V$ a $(1,0)-$tensor)