First of all, I know the fact that exists at least three objects which can be view as the same depending on the context and usage. But suppose then the following:
I) General Points
Given a connection $\nabla_{(\cdot)}(\cdot)$ on a manifold, then we can explore the notion of Parallel Transport. Given the realization of the geometrical meaning of a parallel transport we can define a notion of a direction derivative $\nabla_{\bf{X}}\bf{Y}$.
Ocurred to me some points, though:
$1)$ In a coordinate chart we can write the directional derivarive as:
$$ \nabla_{\textbf{X}}\textbf{Y} = \Big\{X^{\nu}\Big[ \partial_{\nu}Y^{\alpha} + \Gamma^{\alpha}_{\hspace{1.5mm}\mu\hspace{0.2mm}\nu}Y^{\mu}\Big]\Big\}\vec{\partial}_{\alpha} \equiv \Big(X^{\nu}\nabla_{\nu}Y^{\alpha}\Big)\vec{\partial}_{\alpha} \tag{1}$$
$2)$ From General Relativity books and (some) Differential Geometry literature we call Covariant Derivative the components:
$$ \nabla_{\nu}Y^{\alpha} = \partial_{\nu}Y^{\alpha} + \Gamma^{\alpha}_{\hspace{1.5mm}\mu\hspace{0.2mm}\nu}Y^{\mu} \tag{2}$$
We note that this so-called "covariant derivative" is no more than a part of the components directional derivative $(1)$, i.e, is just $\Big(X^{\nu}\nabla_{\nu}Y^{\alpha}\Big)$ without the vector components $X^{\nu}$.
$3)$ From a simple comparison we can speculate that the whole symbol $\nabla_{\nu}Y^{\alpha}$ could given rise to another object such as:
$$\nabla\textbf{Y} = \nabla_{\nu}Y^{\alpha} (some\hspace{1.5mm} tensor\hspace{1.5mm} basis)\tag{3}$$
$4)$ From elementary tensor calculus, which is famously a non-rigorous subject, we can prove that such symbols $\nabla_{\nu}Y^{\alpha}$ indeed transforms as a tensor field:
$$\nabla_{\nu '}Y^{\alpha '} = \frac{\partial x^{\alpha '}}{\partial x^{\beta}}\frac{\partial x^{\gamma}}{\partial x^{\nu '}}\nabla_{\gamma}Y^{\beta} \tag{4}$$
We note that $4)$ is a $(1,1)-$tensor field, in fact. Now, a $(1,1)-$tensor is an object that lives in a tensor product $T_{p}M \otimes T_{p}M^{*}$, where $T_{p}M$ and $T_{p}M^{*}$ are the tangent and cotangent spaces, and therefore is an object just like:
$$ \mathrm{T} = t^{\mu}\hspace{0.3mm}_{\nu} \vec{\partial}_{\mu} \otimes dx^{\nu} \tag{5}$$
and it's components transforms as:
$$ t^{\alpha '}\hspace{0.3mm}_{\nu '} = \frac{\partial x^{\alpha '}}{\partial x^{\beta}}\frac{\partial x^{\nu}}{\partial x^{\gamma '}}t^{\beta}\hspace{0.3mm}_{\gamma} \tag{6}$$
Therefore from the arguments presented and the general realization of $(5)$ and $(6)$, we can properly say that the object $\nabla\textbf{Y}$ is written as:
$$\nabla\textbf{Y} = \nabla_{\nu}Y^{\mu} \vec{\partial}_{\mu} \otimes dx^{\nu} \tag{7}$$
II) My Doubt
Regardless the chosen basis, a tensor (more generally a tensor field) is a multilinear function defined by:
$$ \mathrm{T}: T_{p}M \times....\times T_{p}M \times T_{p}M^{*}\times....\times T_{p}M^{*} \to \mathbb{R} \tag{8}$$
Thefore, the object $(5)$ is a object that acts like:
$$ \mathrm{T}: T_{p}M \times T_{p}M^{*} \to \mathbb{R} \tag{9}$$
The problem is that, I'm sure that the object that transforms as $(6)$ is something that acts like $(9)$. But I'm not sure if the covariant derivative is something that acts like $(9)$. Follwing my reasoning I would said yes, but I never see a covariant derivative acting on a vector and spitting a scalar. So:
A covariant derivative is the object $(7)$? In other words, a covariant derivative is a $(1,1)-$ tensor field?