$
\newcommand\dd{\mathrm D}
\newcommand\Tr{\mathop{\mathrm{Tr}}}
$Let $V$ be our (real) $n$-dimensional vector space. Something that must be noted is that divergence—being the trace of the differential—is an intrinsic quantity, but curl is not in the following sense: given any nondegenerate symmetric bilinear form $(v, w) \mapsto v\cdot w$ (henceforth, a metric), we can define a basis-independent symbol $\nabla$ in such a way that $v\cdot\nabla$ is the $v$-directional derivative and $\nabla\cdot f(x)$ is always the one and only divergence. The curl $\nabla\wedge f(x)$, however, is always dependent on the choice of metric. (Since you mention curl as being a bivector field, I am taking this expression as the definition of curl.)
Taking it on faith for the moment that $v\cdot\nabla f(x) = \dd f_x(v)$, what we want to do is extract a linear map $V \to V$ from the curl somehow. As a tensor, the curl acts on $V^{*}\wedge V^*$ in the following way
$$
\langle\nabla\wedge f(x),\, \omega\wedge\eta\rangle = \omega(\nabla)\eta(f(x)) - \eta(\nabla)\omega(f(x)).
$$
Using the raising operator $\sharp : V^{*} \to V$, we can write this as
$$
(\omega^\sharp\cdot\nabla)\eta(f(x)) - (\eta^\sharp\cdot\nabla)\omega(f(x)) = \eta(\dd f_x(\omega^\sharp)) - \omega(\dd f_x(\eta^\sharp)).
$$
In index notation, this reads
$$
\eta_i\partial_jf^ig^{jk}\omega_k - \omega_k\partial_jf^kg^{ji}\eta_i = (\partial_jf^ig^{jk} - \partial_jf^kg^{ji})\omega_k\eta_i.
\tag{$*$}
$$
Here we see the metric ($g$) dependence very explicitly. To get a matrix, we want to raise one of $\omega$ or $\eta$; choosing $\eta$, we write
$$
(\partial_jf^ig^{jk} - \partial_jf^kg^{ji})g_{il}\omega_k\eta^l = (\partial_jf^ig^{jk}g_{il} - \partial_jf^k\delta^j_l)\omega_k\eta^l = (g^{jk}\partial_jf^ig_{il} - \partial_lf^k)\omega_k\eta^l.
$$
The first term is exactly the metric adjoint $\overline{\dd f_x}$ of the differential, defined for any $T : V \to V$ by
$$
v\cdot T(w) = \overline T(v)\cdot w
$$
for all $v, w \in V$. When $V = \mathbb R^n$ and the metric is the standard inner product, the adjoint is precisely the transpose of $T$ when expressed in the standard basis.
So we have that the following map $V \to V$ is equivalent to the curl of $f$ at $x$:
$$
\overline{\dd f_x} - \dd f_x.
$$
In this sense the curl is the "antisymmetric part" of $\dd f_x$.
Returning to the index notation expression, let's forcibly raise $\omega$ as well:
$$
(g^{jk}\partial_jf^ig_{il} - \partial_lf^k)\omega_k\eta^l = (g^{jk}\partial_jf^ig_{il} - \partial_lf^k)g_{km}\omega^m\eta^l = (\delta^j_m\partial_jf^ig_{il} - \partial_lf^kg_{km})\omega^m\eta^l
= (\partial_mf^ig_{il} - \partial_lf^kg_{km})\omega^m\eta^l.
$$
But now we can absorb the metric into $f$! We get
$$
(\partial_mf_l - \partial_lf_m)\omega^m\eta^l.
$$
This is a new operation which takes in a (not necessarily linear) function $f : V \to V^{*}$—in other words, a one-form—and outputs a two-form. We can realize the parenthized expression as
$$
(\dd f_x)^{*} - \dd f_x
$$
where by $({-})^{*}$ we mean the dual of a linear transformation; because $\dd f_x : V \to V^{*}$ it's dual is also $V \to V^{*}$ and the above expression makes sense. This is the exterior derivative.
What exactly is $\nabla$, and how can we see the metric (in)dependence of the curl (divergence)? How can we see more clearly the relationship between the curl and the exterior derivative?
Let $L : V^*\times V \to W$ be linear in the first argument. We define a symbol $\Delta$ which is analogous to $\nabla$ but covector like instead of vector like; we will write its argument on the left instead of the right so that $(v)\Delta$ is a scalar-like differential operator defined by
$$
(v)\Delta f(x) = \dd f_x(v)
$$
for all $v \in V$.
We want to define $L(\Delta; x)$ with $\Delta$ differentiating the $x$-dependence. To this end, let $v \in V$ and $\omega\in V^*$. We consider the following manipulations reasonable:
$$
L(\omega\,(v)\Delta; x) = (v)\Delta L(\omega; x) = \dd[L(\omega; {-})]_x(v).
$$
Contracting $\omega$ and $v$, which we write as $\Tr_{\omega(v)}$, we define
$$
L(\Delta; x) = \Tr_{\omega(v)}\dd[L(\omega; {-})]_x(v).
$$
Call this the derivative of $L$. We already know of two derivatives:
- Let $f : V \to V$ be any function, and define $L(\omega; x) = \omega(f(x))$. Then its derivative is the divergence of $f$, which we write as $(f(\dot x))\dot\Delta$. We use overdots to make it clear that $\Delta$ is differentiating $x$.
- Let $\Psi$ be a differential form, i.e. a function $V \to \mathop\bigwedge V^*$, and let $L(\omega; x) = \omega\wedge\Psi(x)$. Then the derivative is the exterior derivative, which we write as $\Delta\wedge\Psi(x)$.
Let's write out the last one.
$$
\Delta\wedge\Psi(x) = \Tr_{\omega(v)}\dd[\omega\wedge\Psi
]_x(v) = \Tr_{\omega(v)}\omega\wedge\dd\Psi_x(v).
$$
Now using index notation with a basis $e_i$ with dual $e^i$,
$$
\Delta\wedge\Psi(x) = e^i\wedge\partial_i\Psi,
$$
which is the exterior derivative.
Now, given a metric we have the musical isomorphism $\sharp : V^* \cong V$, and we simply define
$$
\nabla = \Delta^\sharp.
$$
Now it's clear why the divergence of $f : V \to V$ is metric-invariant:
$$
\nabla\cdot f(x) = \Delta^\sharp\cdot f(x) = (f(\dot x))\dot\Delta.
$$
In a sense, the metricity of $\nabla$ and the metricity of $\cdot$ "cancel out". We also see why the curl is metric dependent
$$
\nabla\wedge f(x) = \Delta^\sharp\wedge f(x)
$$
and the relationship with the exterior derivative is readily apparent.
Finally, we easily derive that
$$
v\cdot\nabla f(x) = v\cdot\Delta^\sharp f(x) = (v)\Delta f(x) = \dd f_x(v).
$$
