$\newcommand{\Reals}{\mathbf{R}}\newcommand{\dd}{\partial}\newcommand{\Del}{\nabla}$tl; dr The $\Del$ operator (i.e., the gradient operator on functions and the curl and divergence operators on vector fields in $\Reals^{3}$) turns out to be an artifact of the exterior derivative operator acting on differential forms.
At the same level of generality (first-order differential operators depending only on the smooth structure of a manifold, not on a Riemannian metric), there is no operator corresponding to $\Del F$ for a vector field $F$.
In more detail, if $f$ is a smooth, real-valued function on some non-empty open subset $U \subset \Reals^{3}$, define
$$
df = \frac{\dd f}{\dd x}\, dx + \frac{\dd f}{\dd y}\, dy + \frac{\dd f}{\dd z}\, dz.
\tag{1}
$$
If $\alpha = \alpha_{1}\, dx + \alpha_{2}\, dy + \alpha_{3}\, dz$ is a smooth $1$-form in $U$, define
\begin{align*}
d\alpha
&= d\alpha_{1} \wedge dx + d\alpha_{2} \wedge dy + d\alpha_{3} \wedge dz \\
&= \left[\frac{\dd \alpha_{3}}{\dd y} - \frac{\dd \alpha_{2}}{\dd z}\right] dy \wedge dz
+ \left[\frac{\dd \alpha_{1}}{\dd z} - \frac{\dd \alpha_{3}}{\dd x}\right] dz \wedge dx
+ \left[\frac{\dd \alpha_{2}}{\dd x} - \frac{\dd \alpha_{1}}{\dd y}\right] dx \wedge dy.
\tag{2}
\end{align*}
If $\omega = \omega_{1}\, dy \wedge dz + \omega_{2}\, dz \wedge dx + \omega_{3}\, dx \wedge dy$ is a smooth $2$-form in $U$, define
\begin{align*}
d\omega
&= d\omega_{1} \wedge dy \wedge dz + d\omega_{2} \wedge dz \wedge dx + d\omega_{3} \wedge dx \wedge dy \\
&= \left[\frac{\dd \omega_{1}}{\dd x} + \frac{\dd \omega_{2}}{\dd y} + \frac{\dd \omega_{3}}{\dd z}\right] dx \wedge dy \wedge dz.
\tag{3}
\end{align*}
(Exercise: Use (1) and anti-symmetry of the wedge product to deduce (2) and (3).) In a sense that should be formally apparent, (1) corresponds to the gradient $\Del f$ of $f$, (2) corresponds to the curl $\Del \times F$ of the vector field $F = (\alpha_{1}, \alpha_{2}, \alpha_{3})$, and (3) corresponds to the divergence $\Del \cdot F$ of the vector field $F = (\omega_{1}, \omega_{2}, \omega_{3})$.
It turns out that these operators on differential forms are natural, while the classical vector operators are...not, really (and secretly they rely on extra structure, the Euclidean metric, in their definitions).
From the perspective pf physics or multivariable calculus, the $\Del$ operator should be viewed as a mnemonic, a formal way of remembering how to compute grad, curl, and div.
Finally, assuming I understand your intent, you're expecting something like
$$
\Del F = (\Del F_{1}, \Del F_{2}, \Del F_{3}),
$$
which would formally be a vector field whose components are vector fields, or a matrix-valued function. In this sense, for the example $F = (x, y, z)$ you give, you'd have
$$
\Del F = \left[
\begin{array}{@{}ccc@{}}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{array}\right].
$$
