Differential of $0$-form If I have a $1$-form that in local coordinates can be written as $\alpha=\alpha_i dx^i$ and a vector field $X=\xi^j \frac{\partial}{\partial x^j}$, if I apply $\alpha$ to $X$ I should obtain the following:
$$\alpha(X)=\alpha_idx^i\Bigl(\xi^j \frac{\partial}{\partial x^j}\Bigr)=\color{red}{\alpha_i\xi^j dx^i\Bigl(\frac{\partial}{\partial x^j}\Bigr)}=\alpha_i\xi^j\delta^j_i=\alpha_i\xi^i$$
$\textbf{My question:}$ I can't understand the red expression, in particular why $\alpha_i dx^i\bigl(\xi^j\frac{\partial}{\partial x^j}\bigr)=\alpha_i\xi^j dx^i\bigl(\frac{\partial}{\partial x^j}\bigr)$; why can I put outside the $\xi^j$ (that is a $0$-form) outside the differential as it is a constant?
 A: If $\alpha = \alpha_i \mathrm{d}x^i$ is a $1$-form and $X = X^j \partial_j$ is a vector field, the notation $\alpha(X)$ has to be understood as a function on $M$ defined by
$$\alpha(X)(p) = \alpha_p (X_p).$$
Now, for a fixed point $p \in M$, $\alpha_p$ is a linear form on $T_pM$ and $X_p = X^j(p) \partial_j(p)$ is a vector. Hence, by linearity
$$
\alpha(X)(p) = \alpha_i(p) \mathrm{d}x^i|_p\left(X^j(p)\partial_j(p) \right) = \alpha_i(p)X^j(p)\mathrm{d}x^i|_p \left(\partial_j(p) \right) = \alpha_i(p)X^j(p) \delta^i_j = \alpha_i(p)X^i(p).
$$
And this tells us that $\alpha(X)$ is the function on $M$
$$
\alpha(X) : p \mapsto \alpha_i(p)X^i(p).
$$
Edit Forget about the manifold. Suppose you have a vector space $V$ with basis $(e_1,\ldots,e_n)$ and dual basis of $V^*$ $(e^1,\ldots,e^n)$. Then if $X = X^i e_i \in V$ is a vector and if $\alpha = \alpha_j e^j \in V^*$ is a linear form on $V$, then by definition of the dual basis
$$
\alpha(X) = \alpha_iX^i.
$$
Now, remember that a tangent bundle $TM$ is a familly of vector spaces $(T_pM)_{p\in M}$. Do not think about any differentiable structure or anything. A vector field is a familly $(X_p)_{p\in M}$ with $X_p \in T_pM$. A $1$-form is a familly of covectors $(\alpha_p)_{p\in M}$ with $\alpha_p \in T_pM^*$.
Fix $p\in M$, apply the above reasoning to $V = T_pM$, $X = X_p$ and $\alpha = \alpha_p$.
Comment You seem to be confused about the differential. But the in the expression $\mathrm{d}x^i (X^j\partial_j)$, you are not taking any differential. You are just applying the $1$-form "$\mathrm{d}x^i$" to the vector $"X^j \partial_j"$.
