At the moment I have some trouble with the differential of a function $df$ as covector field. Most of the explanations are on manifolds but I am at the moment interested in $\mathcal{R}^n$.
The function $f$ is defined as $f : \mathcal{R}^n \rightarrow \mathcal{R}$. As I understand it we have at every point $ x \in \mathcal{R}^n $ a covector namely $df(x)$. But now suppose we let $df$ act on a vector field $v=v^i \partial_i \in \mathcal{R}^n$ (in $\mathcal{R}^n$ since at every point $x$ we have a basis for the tangent space $T_x \mathcal{R}^n$ isomorphic to $\mathcal{R}^n$ so no explicit need for tangent spaces in case of $\mathcal{R}^n$) such that $df(v)=v^i \partial_i f$ which can be evaluated for an arbitrary point. My question is: Do these coefficients $v^i$ of the vector field $v$ depend on the coordinates you are working with so $x^1,...,x^n$?
Absolutely! A vector field is an assignment of a tangent vector to each x in $R^n$ of a vector v with coordinates $v^i(x)$ ,i=1,2,...n relative to the standard basis for R^n . I am now responding to your question about the above answer . For a metric tensor one is assigning to each x ,an inner product on the tangent vectors $ u =(u_1...u_n)$ ,$v=(v_1,...v_n)$ ,that is $g(x)=g_{i,j}(x)u_iv_j$ . so $g(i,j)$ varies with x , but u and v are fixed vectors . Of course you could make u and v functions of x also if you wanted to look at the metric tensor along vector fields .Hope this helps . Note we are using the summation conventions ,i.e. we are summing over i,j in the formula for g .