Tensor Components

I would like to ask something

On the Barrett Oneill's Semi-Riemann Geometry there is a definition of tensor component: Let $\xi=(x^1,\dots ,x^n)$ be a coordinate system on $\upsilon\subset M$. If $A \in \mathscr I^r_s (M)$ the components of $A$ relative to $\xi$ are the real-valued functions $A _j^i, \dots j^i =A(dx^i_1,\dots ,dx^i_s, \delta_{j 1},\dots, \delta_{j s})$ (I couldn't write it properly but $i$'s are from $1$ to $r$ and $j$'s are $1$ to $s$) on $\upsilon$ where all indices run from $1$ to $n=dim M$.

(and here is the thing that I didn't get)
By the definition above the $i$th component of $X$ relative to $\xi$ is $X(dx^i)$,which is interpreted as $dx^i(X)=X(x^i)$.

While one-forms are $(0,1)$ tensors we could interprete them like $V(\theta)=\theta(V)$
So we can do the same thing here: $X(dx^i)=dx^i(X)$.
But how did we write $dx^i(X)=X(x^i)$

Do I think something wrong?
(Sorry if I wrote it bad but I couldn't find how to write.)

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The tensor $A_p$ at $p\in M$ of type $(r,s)$ is the multilinear map

$$A_p: \underbrace{T_p^{*} M\times\dots T_p^{*} M}_{r~\text{times}} \times\underbrace{T_p M\times\dots T_p M}_{s~\text{times}}\rightarrow \mathbb R$$

whose components ${A_p}^{i_1\dots i_r}_{j_1\dots j_s}$ are given by

$${A_p}^{i_1\dots i_r}_{j_1\dots j_s}:=A_p(dx^{i_1},\dots,dx^{i_r},\frac{\partial}{\partial x_{j_1}},\dots, \frac{\partial}{\partial x_{j_s}}),$$

denoting by $\xi=(x_1,\dots,x_n)$ a local coordinate system at $p$, and by $\{dx^{\bullet}\}$, $\{\frac{\partial}{\partial x_{\bullet}}\}$ dual basis of $T_p^{*} M$, $T_p M$.

As stated,

$$A_p^{i}:=A_p(dx^i),$$ $${A_p}_j:=A_p(\frac{\partial}{\partial x_j}).$$

The above convenstions and definitions are those which are currently used in most textbooks. I would propose then to stop here and use them for your computations; what comes under "which is interpreted...", is in my opinion not clear (for example, what is $A(x_i)?$ Maybe $A$ evaluated at $(x_1,\dots,x_n)$? But then why to use that notation, if it is what has been using since the beginning of the OP? Etc...).

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The is a very nice intuitive explanation of this in Penrose's Road to Reality, ch 14. As a quick summary, any vector field can be thought of as a directional derivative operator on scalar valued functions, i.e for every scalar valued smooth function $f$ and vector field $X$, define the scalar field $$X(f) \triangleq p \mapsto \lim_{\epsilon \rightarrow 0} \frac{f(p+\epsilon X_p) - f(p)}\epsilon$$ at every point $p$. This action on all $f$ uniquely determines $X$.

Similarly the 1-form $df$ acts on a vector field $X$ to yield the linear change in $f$ along $X$, ie: $$df(X) = X(f)$$

Substituting $x^i$ for $f$ then gives $dx^i(X) = X(x^i)$.

In more detail, given a coordinate system $x^i$, exterior calculus says $$df = \sum_i \partial_{x^i}f\,dx^i$$ Combining the above $$X(f) = df(X) = \sum_i \partial_{x^i}f\,dx^i(X) = \sum_i \partial_{x^i}f \, X(x^i) = \sum_i X(x^i) \, \partial_{x^i} f \\ \therefore X = \sum_i X^i \partial_{x^i} \text{ where } X^i \triangleq X(x^i)$$

So $\partial_{x^i}$ form a basis for the space of directional derivative operators and thus also vector fields.

The $dx^i$ then are the dual basis of $\partial_{x^i}$, since $$dx^i(\partial_{x^j}) = \partial_{x^j} x^i = \delta^i_j \\ dx^i(X) = X(x^i) = X^i$$ as required.

To formalize this for the general Riemannian setting, you need express it in terms of manifolds, charts and tangent spaces/bundles, with appropriate smoothness conditions on everything, but Penrose's explanation gives you a nice mental picture to start with. The book also has excellent diagrams. It is worth getting just for differential geometry chapter.

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