Let $U \subseteq \mathbb{R}^n$ be open. I first want to show/acknowledge that the mapping
$$\omega_p(v_1, ..., v_{n-1}) := det(p, v_1, ..., v_{n-1}), p \in U, v_i \in \mathbb{R}^n$$
defines a differential form of degree $n-1$, and then want to find it's "base representation", i.e. a representation of $\omega$ of the form
$$\omega = \sum_{1 ≤ i_1 < ... < i_k ≤ n} a_{i_1 ... i_k} d x_{i_1} \wedge ... \wedge d x_{i_k} $$
where the $a_{i_1 ... i_k}: U \to \mathbb{R}$ are coefficient functions, and where the $d x_{i}$ are the differentials of the canonical coordinate functions on $U$, i.e. $(d x_i)_p(v) = v_i$ for any $v \in U$, where $v_i$ is the $i$-th coordinate of $v$.
Now, regarding the first point: per my definition, $\omega$ is a differential form if $\omega_p \in \Lambda^k(\mathbb{R}^n)^*$ (the set of alternating $k$-forms over $\mathbb{R}^n$) for any $p \in U$. But I already know that the determinant for an $n \times n$-matrix is an alternating $n$-form, and since for any $p \in U$, the $p$ is "fixed" in the expression $det(p, v_1, ..., v_{n-1})$, it should follow right away that $\omega$ is indeed an $n-1$-form? (Maybe my phrasing of this argument isn't the most elegant, but is my reasoning here correct and it's really that simple, or is it more complicated?)
I don't really know how to approach the second point and find such a base representation. I know that such a base representation always exists, but how can I find the right $i_1, ..., i_k$ and the right $a_{i_1 ... i_k}$'s? Differential forms are still very new to me, so it would be awesome if someone could help me out.
Let $x=(x_1,...,x_n)$ be (global) coordinates for $\mathbb{R}^n$. Then $p$ lives in the tangent space, $T\mathbb{R}^n|_x$, which has corresponding basis
$\dfrac{\partial}{\partial x_1},...,\dfrac{\partial}{\partial x_n}$
The dual space to the tangent space, $T^{\star}\mathbb{R}^n$ is the cotangent space, which is the space of $1$-forms. It has the corresponding basis
$dx_1,...,dx_n$
A $1$-form takes a vector field as input, and one has that $\langle dx_i;\partial/\partial x_j\rangle=\delta_j^i$ where $\delta_j^i$ is the Kronecker delta ($1$ if $i=j$, $0$ otherwise).
To proceed to higher forms, one needs the wedge product, which is defined by the determinantal formula
$\langle\omega_1\wedge...\wedge\omega_k;v_1,...,v_k\rangle=\det\langle\omega_i;v_j\rangle$
which is why it's an alternating $k$-linear map.
The interior product is an antiderivation on forms, taking $k$-forms to $(k-1)$-forms. It is defined by
$\langle v\lrcorner\omega;v_1,...,v_{k-1}\rangle=\langle\omega;v,v_1,...,v_{k-1}\rangle$
This matches the situation with $p$ in your problem. Note that $p$ and the $v_i$'s are, most generally, vector fields, which evaluate to individual vectors once you specify a point on your manifold at which to evaluate them. Hence they have coordinate expressions of the form
$p(x)=p_1(x)\partial/\partial x_1+...+p_n(x)\partial/\partial x_n$
and similarly for each $v_i$ (bewary of index confusion in how you choose to notate the individual coordinate functions for these!).
A $1$-form, $\omega$ can be expressed in the given coordinates as
$\omega_1(x) dx_1+...+\omega_n(x) dx_n$
A $2$-form $\theta$ has the coordinate expression
$\underset{i< j\in\{1,...,n\}}{\sum}\theta_{i,j}(x)dx_i\wedge dx_j$
I think you can infer what $k$-form looks like from this.
In your first point, your reasoning is roughly correct, you have an $n$-form which has one of the input vector fields, hence you effectively have one less degree of freedom for the resulting form, and this is exactly what an interior product does, takes a $k$-form to a $(k-1)$-form.
$n$-forms on an $n$-dimensional manifold have a nice expression in local coordinates (which are actually global coordinates in our case).
$\omega (x)dx_1\wedge...\wedge dx_n$
Where I've shifted the index on the vector fields so it will match the situation after we begin working on the interior product with $p$.
Tying all of this together, for the second part what happens is you first apply the determinantal formula to get an expression in terms of a wedge product of $dx_i$'s, you then need to simplify the resulting expression. To do that, you need to substitute in the coordinate expressions for each vector field and then distribute over the sums (which you can do by linearity). You end up with terms that look like:
$(v_i)_j(x)\omega(x)\partial/\partial x_j dx_1\wedge...\wedge dx_n$
where $(v_i)_j(x)$ denotes the $j$-th coordinate function of $v_i$. The last step will be to get rid of the partials which is the evaluation of the interior product. The partial acts by the following rule:
$\partial/\partial x_j dx_1\wedge...\wedge dx_n=(-1)^{i-1}dx_1\wedge...\wedge dx_{i-1}\wedge dx_{i+1}\wedge...\wedge dx_n$
if $dx_i$ is in the wedge product, $0$ otherwise.