A $(k,l)$-tensor field $A$ on smooth manifold $M$ is a smooth section
$$
A : M \longrightarrow T^{(k,l)}TM,
$$
where $T^{(k,l)}TM = \coprod_{p \in M} \underbrace{T_pM \otimes \cdots \otimes T_pM}_{k \text{ times}} \otimes \underbrace{T^*_pM \otimes \cdots \otimes T^*_pM}_{l \text{ times}} $.
Any tensor field has property that for any smooth vector fields $X_1,\dots,X_l$ and smooth covector fields $\omega^1,\dots,\omega^k$ we have a smooth function
$$
A(\omega^1,\dots,\omega^k,X_1,\dots,X_l) : M \longrightarrow \mathbb{R}
$$
defined as $A(\omega^1,\dots,\omega^k,X_1,\dots,X_l)(p) = A_p(\omega^1|_p,\dots,\omega^k|_p,X_1|_p,\dots,X_l|_p) \in \mathbb{R}$. So $(k,l)$-tensor field $A$ induces a map
$$
\tilde{A} : \underbrace{\mathfrak{X}^*(M)\times\cdots\times\mathfrak{X}^*(M)}_{k \text{ times}} \times \underbrace{\mathfrak{X}(M) \times \cdots \times \mathfrak{X}(M)}_{l \text{ times}} \longrightarrow C^{\infty}(M).
$$
An important fact is that this map is multilinear over $C^{\infty}(M)$. It turns out that this in fact characterize tensor fields.
$\textbf{Tensor Characterization Lemma}$ ( Lee's Introduction to Smooth Manifolds ) A map
\begin{equation}
\tau : \underbrace{\mathfrak{X}^*(M)\times\cdots\times\mathfrak{X}^*(M)}_{k \text{ times}} \times \underbrace{\mathfrak{X}(M) \times \cdots \times \mathfrak{X}(M)}_{l \text{ times}} \longrightarrow C^{\infty}(M) \qquad \color{blue}{(\star)}
\end{equation}
is induced by a smooth $(k,l)$ tensor field as above iff this map is multilinear over $C^{\infty}(M)$.
In the proof of the above lemma, the tensor induce from the multilinear map $\tau : \mathfrak{X}^*(M) \times \cdots \mathfrak{X}^*(M) \times \mathfrak{X}(M)\times \cdots \mathfrak{X}(M) \longrightarrow C^{\infty}(M)$ is a tensor field $A : M \to T^{(k,l)}TM$ defined as
$$
A_p(w^1,\dots,w^k,v_1,\dots,v_l):= \tau(\omega^1,\dots,\omega^k,X_1,\dots,X_l)(p)
$$
where $\omega^i\in\mathfrak{X}^*(M)$ is any smooth extension of $w^i\in T^*_pM$ and $X_i \in \mathfrak{X}(M)$ is any smooth extension of $v_i \in T_pM$, for each $i$ (the map $A$ independent of the extensions).
Because of this lemma, we often identify $A$ with its induced map $\tilde{A}$.
In your case above the map is
\begin{equation}
R : \mathfrak{X}(M)^* \times \mathfrak{X}(M) \times \mathfrak{X}(M) \times \mathfrak{X}(M) \rightarrow C^{\infty}(M)
\end{equation}
defined as $$R(\omega,X,Y,Z) := \omega(\nabla_X\nabla_Y Z - \nabla_Y\nabla_X Z - \nabla_{[X,Y]}Z)$$ is defines a $(3,1)$-tensor field if you can show that $R$ is linear over $C^{\infty}(M)$ on each arguments. This including the first and the fourth argument which is not mentioned before in the comments above. The first argument is obvious
$$
R(f\omega,\cdot,\cdot,\cdot) = (f\omega)(\cdot) = f\,\omega(\cdot)
$$
The second and the third and the fourth is similar, it follows from
\begin{align*}
\nabla_{fX_1+gX_2}&= f\nabla_{X_1} + g \nabla_{X_2} \quad \textbf{(Linearity)}\\
\nabla_X(fY) &= f\nabla_X Y + (Xf)Y \quad \textbf{(Leibniz Rule)}\\
[fX,Y] &= f[X,Y] -(Yf)X
\end{align*}
For example,
\begin{align*}
R(\omega,fX,Y,Z) &= \omega (\nabla_{fX} \nabla_Y Z - \nabla_Y \nabla_{fX} Z-\nabla_{[fX,Y]}Z) \\
&= \omega (f\nabla_X \nabla_Y Z - \nabla_Y (f\nabla_XZ ) - \nabla_{f[X,Y]-(Yf)X}Z) \\
&=\omega (f\nabla_X \nabla_Y Z - f\nabla_Y\nabla_XZ - \require{cancel}{\cancel{(Yf)\nabla_XZ}} - f\nabla_{[X,Y]} Z + {\cancel{(Yf)\nabla_XZ}}) \\
&= f R(\omega,X,Y,Z).
\end{align*}
This lemma is oftenly used in Riemannian geometry text implicitly. You can see the proof in Lee's book here p.318.
Beside the type of maps in $\color{blue}{(\star)}$ above, another but similar way to spot a disguised tensor field is through these kind of maps
$$
\tau_0 : \underbrace{\mathfrak{X}^*(M)\times\cdots\times\mathfrak{X}^*(M)}_{k \text{ times}} \times \underbrace{\mathfrak{X}(M) \times \cdots \times \mathfrak{X}(M)}_{l \text{ times}} \longrightarrow \color{red}{\mathfrak{X}(M)}.
$$
If this map is multilinear over $C^{\infty}(M)$ then we can define a map
$$
\tau : \underbrace{\mathfrak{X}^*(M)\times\cdots\times\mathfrak{X}^*(M)}_{\color{red}{k+1 \text{ times}}} \times \underbrace{\mathfrak{X}(M) \times \cdots \times \mathfrak{X}(M)}_{l \text{ times}} \longrightarrow C^{\infty}(M)
$$
as $\tau(\omega^{1},\dots,\omega^{k}, \omega^{k+1},X_1,\dots,X_l) = \tau_0(\omega^{1},\dots,\omega^{k},X_1,\dots,X_l)(\omega^{k+1})$, which is also multilinear over $C^{\infty}(M)$. Therefore $\tau_0$ induced by a smooth $(k+1,l)$-tensor field (as its induced map $\tau$ induce them in the Tensor Characterization Lemma above) iff $\tau_0$ is multilinear over $C^{\infty}(M)$. Other respective variants of $\tau_0$ can be deduced the same way.
Best Answer
Addressing your question about $d\omega$ and $\omega\wedge\omega$:
If the trivial bundle is $E = M \times \mathbb{R}^k$, then each standard basis vector $e_j \in \mathbb{R}^k$ defines a section of $E$ and therefore, there is a a matrix of $1$-forms $\omega^i_j$ such that $$\nabla_X e_j = \langle\omega^i_j,X\rangle e_i.$$ Denote that matrix by $\omega$. Therefore, given a section $s = s^je_j$, $$\nabla s = ds^j e_j + s^j \nabla e_j = ds^j e_j + s^j\omega_j^ie_i = (ds^j + s^i\omega^j_i)e_j = (d + \omega)s.$$ Then $d\omega$ is simply the matrix whose components are $d\omega^i_j$ and $\omega\wedge\omega$ is the matrix of 2-forms whose components are $\omega^i_p\wedge\omega^p_j$.
Here's another way to say the same thing: First, note that treating $E$ as a trivial bundle is essentially the same as choosing a frame of sections of $E$ and writing any section of $E$ with respect to the frame.
In particular, for each $X \in T_pM$, there is a matrix $A_X$ of coefficients such that $$ \nabla_X s_j = s_i(A_X)_j^i. $$ Since the map $X \rightarrow A_X$ is point wise linear, there exist $1$-forms $\omega^i_j$ such that $$ \nabla s_j = s_i\omega_j^i. $$ Given any section $s = a^js_j$, $$ \nabla s = \nabla(a^js_j) = da^js_j + a^j\omega^i_js_i = (da^j + a^i\omega_i^j)s_j $$ In this sense, $$ \nabla = d + \omega. $$
The curvature tensor at $p \in T_pM$ is defined as follows: Given $X, Y\in T_pM$ and a section $s$ of $E$, it is $$ F(X,Y)s = \nabla_X(\nabla_Ys) - \nabla_Y(\nabla_Xs) - \nabla_{[X,Y]}s \in E_p. $$ You can verify that $F(X,Y)s$ depends only on $s(p)$ and therefore for each $p \in M$ and $X, Y \in T_pM$, defines a bundle map $$ F(X,Y): E_p \rightarrow E_p. $$ In particular, for each $p \in M$ and $X, Y \in T_pM$, there exists a matrix $B_{X,Y}$ such that $$ F(X,Y)s_j(p) = s_i(p)(B_{X,Y})^i_j. $$ It can be verified that each $(B_{X,Y})^i_j$ defines a differential $2$-form $\Omega^i_j$ on $M$ such that $$ (B_{X,Y})^i_j = \langle \Omega^i_j, X\otimes Y\rangle. $$ Therefore, $$ F(X,Y)s_j = s_i\langle\Omega^i_j,X\otimes Y\rangle. $$ On the other hand, \begin{align*} F(X,Y)s_j &= \nabla_X(\nabla_Ys_j) - \nabla_Y(\nabla_XSs_j) - \nabla_{[X,Y]}s_j\\ &=\nabla_X(s_i\langle\omega^i_j,Y\rangle) - \nabla_Y(s_i\langle\omega^i_j,X\rangle) - s_i\langle \omega^i_j,[X,Y]\rangle\\ &= (\nabla_Xs_i)\langle\omega^i_j,Y\rangle + s_i(X\langle\omega^i_j,Y\rangle) - (\nabla_Ys_i)\langle\omega^i_j,X\rangle - s_i(Y\langle\omega^i_j,X\rangle) - s_i\langle\omega^i_j,[X,Y]\rangle\\ &= s_i(\langle \omega^i_k,X\rangle\langle\omega^k_j,Y\rangle + X\langle\omega^i_j,Y\rangle - \langle \omega^i_k,Y\rangle\langle\omega^k_j,X\rangle - Y\langle\omega^i_j,X\rangle - \langle \omega^i_j,[X,Y]\rangle)\\ &= s_i(X\langle\omega^i_j,Y\rangle - Y\langle \omega^i_j,X\rangle - \langle \omega^i_j,[X,Y]\rangle + \langle\omega_k^i,X\rangle\langle\omega^k_j,Y\rangle - \langle \omega_k^i,Y\rangle\langle\omega^k_j,X\rangle\\ &= S_i(d\omega^i_j + \omega^i_k\wedge\omega^k_j). \end{align*} From all this, follows that $$ F(X,Y)s_j = s_i(d\omega^i_j + \omega^i_k\wedge\omega^k_j), $$ i.e., $$ \Omega^i_j = d\omega^i_j + \omega^i_k\wedge\omega^k_j $$