How to get the linear combination of a k-form?

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Let M be a smooth manifold then a k-form on it is defined as follows:

$w: M \rightarrow \Lambda^k T_p M \\ p \mapsto w_p$

Where $w_p$ is a k-form on the tangent space. We call such an element differentiable if and only if for all k-tuples $(v_1,...,v_k)$ with $v_i \in T_p M$ the transformation

$w_p: T_pM \rightarrow \mathbb{R}$

is differentiable for every such k-tupel.

Further we call the the elements $w_{i_1...i_k}:=w(e_{i_1},...,e_{i_k}): U \rightarrow \mathbb{R}$ the compnentfunctions of $w$ with a basis B of $\mathbb{R}^n$ so that $B=(e_1,...,e_n)$.

Then we might write $w$ as follows:

$w=\sum_{i_1<...<i_k} w_{i_1...i_k} dx^{i_1}\wedge...\wedge dx^{i_k}$

with the componentfunctions as defined above and $dx^{i_1}\wedge...\wedge dx^{i_k}$ as elements of the basis of $\Lambda^kT_pM$.

I know that we get every k-form as linear combination of the elements I wedged there together with the $w_{i_1...i_k}$ being scalars.

My question is now: Why are the scalars now that we look at the k-form on a smooth manifold functions and not scalars?

My hypothesis is that we look at $w_{i_1...i_k}$ as function of p, thus to say that we locked some argument values $(v_1,...,v_k)$ and do only insert p.