With the euclidean metric I use the musical isomorphisms to obtain $1$-form associated with a vector field, so for a vector field $\vec{F}=(f_1,f_2,f_3)$ we have $ \vec{F}^{\flat}=f_1dx+f_2dy+f_3dz$ and $2$-form associated with $F$ is $f_1 dy\wedge dz+f_2 dz\wedge dx+f_3 dx\wedge dy$, how can I do the exactly same thing with a general metric?
Thanks
On
Suppose $(M,g)$ is an $n$-dimensional Riemannian (or pseudo-Riemannian) manifold. The metric $g$ determines an isomorphism $\flat\colon TM\to T^*M$, defined for any vector field $X$ by $$ X^\flat = g(X,\cdot), $$ or equivalently $$ X^\flat(Y) = g(X,Y) \qquad \text{for every vector field $Y$}. $$
(This is the vector bundle version of the map defined by @JonHerman.) Thus every vector field determines a unique $1$-form.
If $M$ has an orientation, $g$ also determines an isomorphism $\beta$ from $TM$ to the bundle of $(n-1)$-forms by $$ \beta(X) = i_X dV, $$ where $dV$ is the Riemannian volume form and $i_X$ denotes interior multiplication. More explicitly, this means $$ \beta(X) (Y_1,\dots,Y_{n-1}) = dV(X,Y_1,\dots,Y_n) \qquad \text{for vector fields $Y_1,\dots,Y_{n-1}$}. $$
In the special case that $M$ is $\mathbb R^3$ with the Euclidean metric, these two maps are the ones you wrote down.
On a finite-dimensional vector space $V$, a metric is a symmetric positive definite map $g:V\times V\to \mathbb{R}$. Any non-degenerate bilinear map on a finite dimensional vector space $V$ sets up an isomorphism between $V$ and $V^\ast$, the dual space of $V$. This is done through the map $$V\to V^\ast \ \ \ \ \ \ \ \ \ \ v\mapsto g(v,\cdot)$$ The non-degeneracy shows this map is injective, while we know $V\cong V^\ast$. A metric is positive definite, so in particular non-degenerate, and so we get this correspondence.
Also, in your example the vector field $F$ in $T_pM$ is associated to the $1$-form $F^\flat$ in $T_p^\ast M$. The musical isomorphism doesn't also associate a $2$-form to $F$.