Let $V \in T_p M$ be a vector at the point $p$ in a manifold $M$. The connection $\nabla$ is defined such that $\nabla V$ is a $(1,1)$ tensor: \begin{align*} \nabla V: T_p M \rightarrow T_p M \otimes T_p^*M . \end{align*} If $X$ be another vector at $p$ then we define the covariant derivative of $V$ in the direction of $X$ as, \begin{align*} \nabla_X V: T_p M \otimes T_p M \rightarrow T_p M. \end{align*} Thus, $\nabla_X V$ is a $(1,0)$ tensor. We also write, \begin{align} (\nabla V) (X) = \nabla_X V \qquad \qquad (1) \end{align} However, this is also written as, \begin{align} \nabla_X V = X \cdot \nabla V = X^a \nabla_{e_a} V \qquad (2) \end{align} where ${e_a}$ is a set of basis vectors. I do not understand the meaning of the last expression in eq. 2. I also have a confusion about what is denoted by the symbol $\nabla$. By itself, as far as I understand, $\nabla$ is not a $(0,1)$ tensor i.e. $\nabla \neq \nabla_a e^a$ (where ${e^a}$ forms a set of basis co vectors) but acting on a vector or, in general, a tensor of rank $(p,q)$ it yields a tensor of rank $(p,q+1)$. Am I correct?
Now, how do we define $Div V$? I read in the post Divergence in terms of Levi-Civita connection that \begin{align*} Div V = trace(\nabla V). \end{align*} To take the trace, I need to write $\nabla V$ as $\nabla_a V^b$, may be in abstract index notation. But I do not understand why we can put the covariant index $a$ on $\nabla$.
P.S.: I am not a mathematician and am trying to learn writing precisely. So, if there are mistakes in my notations, please do point them out.