How to linearly extend the DeRahm operator $d \colon \Omega^{*}(M) \to \Omega^{*+1}(M) $

Question

I have two questions regarding section 1.2.5. of the following lecture notes.

Question 1 (Answered)

In the lecture notes, they view a differential form $\omega \in \Omega^p(M)$ as a $C^{\infty}$-antisymmetric linear map $$ \omega \colon \underbrace{\mathfrak{X}(M) \times \cdots \times \mathfrak{X}(M)}_{p \text{ times}} \to C^{\infty}(M). $$ They then claim that you may write $$ d(\omega)(X_1, \ldots, X_{p+1}) = \sum_{i < j}(-1)^{i + j}\omega([X_i, X_j], X_1, \ldots, \hat X_i, \ldots, \hat X_j, \ldots, X_{p+1}) + \sum_{i = 1}^{p+1} (-1)^{i + 1}L_{X_i}(\omega(X_1, \ldots, \hat X_i, \ldots, X_{p+1})). $$ Here the hat above the vector fields denotes that they leave that entry out. Also $L_X$ denotes the Lie derivative along the vector field $X$.

I don't see how or why this arises in this exact way. Could someone maybe give a hint as to why this is the case?

EDIT: As has been pointed out by Berci, I should clarify which definition I am using for the operator $d$. The definition I am using is that $d \colon \Omega^{\bullet}(M,E) \to \Omega^{\bullet + 1}(M,E)$ is fully characterized by

1. The operator $d \colon \Omega^0(M) = C^{\infty}(M) \to \Omega^1(M)$, is simply the differential, i.e. it is given by $df = \sum_{i = 1}^{n} \frac{d f}{d x_i} d x^i$.
2. The operator satisfies the Leibniz identity, i.e. $d(\omega \wedge \eta) = d (\omega) \wedge \eta + (-1)^{|\omega|} \omega \wedge d(\eta)$
3. $d \circ d = 0$.

Apparently, this description gives rise to a unique operator. I was under the impression that this wasn't the case for some reason. I will review this later. Thanks @Berci!

Question 2

Using this description for differential forms, they also use this description for $E$-valued differential forms $\omega \in \Omega^p(M, E)$ $$ \omega \colon \underbrace{\mathfrak{X}(M) \times \cdots \times \mathfrak{X}(M)}_{p \text{ times}} \to \Gamma(E). $$ They do this by replacing the Lie derivative with the connection, giving us the operator $$ d_\nabla \colon \Omega^{\bullet}(M,E) \to \Omega^{\bullet + 1}(M,E) $$ This makes sense to me. However, the author goes on and states

Moreoever, $d_{\nabla}$ defined on $ \Omega^{\bullet}(M,E)$ is uniquely determined by what it does on the degree zero part and the fact that $d_{\nabla}$ satisfies the Leibniz identity.

I have tried proving this but to no avail. I have tried searching on StackExchange and Wikipedia for some hints, and the best that I could find was on this Wikipedia page under "exterior derivative". There it is stated that it is due to linearity (and the Leibniz identity). What is exactly meant by that? Do they mean that I should view the space of linear operators $\Omega^{\bullet}(M,E) \to \Omega^{\bullet + 1}(M,E)$ (obeying Leibniz) obeying as a sort of vector space that has the property that its basis elements are the $d \colon \Omega^0(M,E) = \Gamma(E) \to \Omega^1(M,E)$ (which are exactly the connections on $E$)?

Thank you in advance for the hints and the help! :)