I asked this question recently (Basis of differential one-form confusion), thought I understood the answer, but now realise I don't. Lee (Introduction to Smooth Manifolds) says that at a point $p$ and with a vector field $X$ we define a covector field $df$, called the differential of $f$, by$$df_{p}\left(X_{p}\right)=X_{p}f.$$
My original question was, is that definition true for any basis or only for a coordinate basis? I was told that because of the absence of any indices in this definition it was coordinate-independent. OK, I then asked for an example with indices using a non-coordinate basis. @Phoenix87 then responded with:
“choose a basis $\{e_{k}\}$ for the tangent space at $p$, then $X_{p}=a^{k}e_{k}$ and therefore $df(X)|_{p}=a^{k}e_{k}(f)$.”
I thought I understood that, but now I'm confused as to why there's a basis vector on the rhs. As I understand things (tenuous at best, I admit), a one-form acts on a vector at a point to spit out a number. This happens because if $\omega^{a}$ is the one-form basis and $e_{k}$ is the vector basis then, again by definition, $$\omega^{a}e_{k}=\delta_{k}^{a}.$$
So why is there an orphan $e_{k}$ on the rhs of @Phoenix87's answer? Also, what is the non-coordinate one-form basis of $df$ on the lhs?
I know this is really basic stuff, but I just don't get it. I'm a self-studier, by the way.
The $e_k$ is not an orphan and the fact that you think it is, makes me suspect that you miss the summation convention. Under that convention, when an expression contains an index both as an upper index and as a lower index, the expression is interpreted as a sum over that index. Restating more explicitly:
Choose a basis $\{e_1,\ldots,e_n\}$ for the tangent space at $p,$ then for every vector $X_p$ in that space there exist unique coefficients $a^1,\ldots,a^n$ such that
$$X_p=\sum_{k=1}^na^ke_k$$
Covectors are elements of the dual vector space, i.e, linear maps from the original vector space to the real numbers. The covector $df$ at the point $p$ is given by its effect on a general vector $X_p$
$$df(X_p)=\sum_{k=1}^na^k(e_k(f)).$$