Connection between the Jacobian of a vector function and exterior derivative of associated differential form. Let $A\subseteq\mathbb{R}^n$ be an open set and
\begin{equation}
F:A\rightarrow R^n \qquad F\in C^1(A)
\end{equation}
a vector field.
The Jacobian matrix is defined as:
\begin{equation}
J(\boldsymbol{x})=\left(\frac{\partial F_i}{\partial x_j}\right)  \quad \forall\boldsymbol{x}\in A
\quad i,j=1...n
\end{equation}
Since we are not dealing with $\mathbb{R^3}$, I shall talk about exterior derivatives rather than curl.
Now, the exterior derivative of a 0-form (scalar function) is the differential, whose representative vector is the gradient of the function.
The exterior derivative of a differential 1-form  $\omega$ is the 2-form $d\omega$ that can be represented through a proper matrix. On the other hand, the Jacobian matrix plays for vector field the same role that the gradient plays for scalar functions, so by analogy with 0-forms, does the Jacobian matrix of the vector field associated with $\omega$ represent $d\omega$?
If it doesn't, which of these two is the differential $dF$ of the  vector field? Does this happen only for 0-forms?
 A: They are not the same thing. For instance, if you take the exterior derivative of a differential of a function, this is always zero (because $d(d\omega)=0$ for every differential form $\omega$, in particular for $0$-forms), while the Jacobian of the gradient of the function is the Hessian of the function, which is in general non-zero. The above is related to the fact that the curl of the gradient of a function is always 0, as you also suggest.
If you want to compute "the differential" of the vector field, it depends on what you mean and what you have to do with that. In general, the Jacobian gives all the possible information on the vector field up to the first order of approximation (that is, up to an $o(\|x\|)$), while the exterior derivative does not (even if it doesn't mean it is less useful in general, they are in principle very different things).
Edit:
If we want to compare $d$ and $J$ as both objects that are squared matrices with inside the derivatives of their arguments, I think the exterior derivative is just the antisymmetric part of the Jacobian. That's why if the field is the differential of a function, then its exterior derivative is $0$ (the Hessian of a function is a symmetric matrix).
Edit 2:
In my opinion, the main fact for which "the differential of a vector field" seems not to exist canonically is that in differential manifolds you don't have in general an intrinsic way to define derivatives of a vector field if you don't specify "with respect to what" you are derivating. You can of course express the derivatives of your vector field in coordinates, but this will not be a tensor, nor any sort of nontrivial intrinsic object (for instance, you can always choose a set of coordinates on your manifold in order to set the derivatives of your vector field to $0$). While instead, you can take the exterior derivative of a differential form also on manifolds, in particular of functions, and this is a well-defined intrinsic object, that behaves well under an arbitrary change of variables.
Thus, in $\mathbb{R}^n$ the name given to the derivatives of a vector field is just "Jacobian", while for differential forms (and functions) we use "differential". Also have n mind that a vector field in differential manifolds is not a 1-form, thus it is often not a good thing to write "$dF$", even if you are in $\mathbb{R}$.
