Tensor Components In 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 $A _j^i, \dots j^i =A(dx^i_1,\dots
> ,dx^i_s, \delta_{j 1},\dots, \delta_{j s})$ $i=1,\dots,r,\ j=1,\dots,s$ on
  $\upsilon$ where all indices run from $1$ to $n=\dim M$.  
By the definition above the $i$th component of $X$ relative to
  $\xi$ is $X(dx^i)$,which is interpreted as $dx^i(X)=X(x^i)$.

I don't understand the last sentence.
Because one-forms are $(0,1)$ tensors we could interpret 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 $dx^i(X)=X(x^i)$? 
Did I make a mistake?

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.
