Can the formula for the exterior derivative be extended to vector valued differential forms? If $\omega$ is a real $k$-form then the exterior derivative $d\omega$ can be expressed as $
\begin{align*}
d\omega(X_0,\dots,X_k)&=\sum_{i=0}^k (-1)^iX_i(\omega(X_0,\dots,\hat{X}_i,\dots,X_k)) \\
&\quad+\sum_{j=1}^k\sum_{i=0}^{j-1}(-1)^{i+j}\omega([X_i,X_j],X_0,\dots,\hat{X}_i,\dots,\hat{X}_j,\dots,X_k)
\end{align*}
$
where $X_0,\dots,X_k$ are vector fields and $\hat{X}_i$ indicates the omission of the argument $X_i$. 
Is there a way to extend this formula to the case when $\omega$ takes values in some finite dimensional real vector space $V$? The problem appears to be defining $X_i(\omega(X_0,\dots,\hat{X}_i,\dots,X_k))$ when $\omega$ is $V$-valued.
 A: Yes. Like you noticed, it is enough to define the exterior derivative $d$ on $V$-valued functions on $M$ (also known as zero forms with values in $V$). So let $f \colon M \rightarrow V$ be a smooth function where $V$ is a finite dimensional (real or complex) vector space with the natural topology and smooth structure induced by the vector space structure. A tangent vector $v \in T_pM$ (considered as an equivalence class $v = [\alpha]$ of curves) acts on $f$ by the same formula that works for scalar functions:
$$ vf|_p = \frac{d}{dt} f(\alpha(t))|_{t = 0} = \lim_{t \to 0} \frac{f(\alpha(t)) - f(p)}{t}.$$
Alternatively, you can set $vf|_p = P_{f(p)}(df|_p(v))$ where $df_p \colon T_pM \rightarrow T_{f(p)} V$ is the standard differential of maps between manifolds and $P_{f(p)} \colon T_{f(p)} V \rightarrow V$ is the natural identification between $T_{f(p)} V$ and $V$ that comes from the vector space structure. 
Much more concretely, by choosing a basis $(e_1,\dots,e_n)$ for $V$, a smooth function $f \colon M \rightarrow V$ is the same as a collection of $n$ smooth functions $f^i \colon M \rightarrow \mathbb{F}$ where $f = f^i e_i$ (summation convention in place). Then $vf|_p = vf^i|_p e_i = df^i|_{p} (v) e_i$.
This gives you a definition of $df$ as a one-form on $M$ with values in $V$ and then $Xf = df(X)$.
A: Yes, let $\omega = \sum_i \omega_iv_i$ be a vector valued form for some basis $v_i$ of $V$, then we can define $d\omega = \sum_i (d\omega_i)v_i$, i.e. you just extend by linearity everywhere.
A: The extension of the exterior derivative to vector- or tensor-valued forms is called the exterior antisymmetric derivative. It is presented in a concise way in Misner & al (1973), chapter 14.
The exterior derivative of a vector-valued form brings up the problem of subtracting two vectors, because this subtraction underlies the notion of derivative. In the most general case we assume that we have a copy of the vector space $V_P$, isomorphic to $V$, at every point $P$ of the manifold. A vector-valued form is then a map
$$P \mapsto v(P)\otimes\omega(P),$$
where $v(P) \in V_P$ and $\omega$ is a form.
What's crucial here is that there needs not be any canonical isomorphism among the vector spaces $V_P$, although all are non-canonically isomorphic to $V$ and to one another.
To extend the exterior derivative we therefore need a way of comparing vectors on different points $P$, that is, an affine connection. The connection introduces a covariant derivative $\nabla$, by which we compare vectors at different points and can thus speak of their derivative.
The exterior antisymmetric derivative $\mathrm{D}$ is then defined as follows:


*

*on a scalar-valued form $\omega$ (that is, an ordinary form) it coincides with the exterior derivative $\mathrm{d}$:
$$\mathrm{D}\omega := \mathrm{d}\omega$$

*on a vector-valued 0-form $v$ (that is, an vector field with values in $V_P$) it coincides with the covariant derivative:
$$\mathrm{D}v :=\nabla v$$

*on the exterior product of a vector-valued $a$-form $S$ and a scalar-valued $b$-form $\omega$ it satisfies the usual antiderivation rule
$$\mathrm{D}(S\land\omega) =
(\mathrm{D}S) \land \omega + 
(-1)^{a} S \land \mathrm{D}\omega$$
These rules define it completely.
This definition also applies when $V$ is considered "outside" the manifold, as you seem to assume. This case is equivalent to saying that the ${V_P}$ form a trivial vector bundle over the manifold, and therefore they automatically have a flat affine connection, and their covariant derivative reduces to the usual one described in @levap's answer.
The exterior antisymmetric derivative has many fascinating properties, and one especially to be kept in mind: applied twice, it isn't always zero, unlike the ordinary exterior derivative: $\mathrm{D}^2 \ne 0$ in some cases. Its square is in fact connected to the Riemann curvature tensor $\pmb{R}{}^\bullet{}_{\bullet\bullet\bullet}$: for a vector-valued 0-form (that is, an ordinary vector field),
$$\mathrm{D}^2 v= \pmb{R}\cdot v$$
the contraction being on the second slot of $\pmb{R}{}^\bullet{}_{\bullet\bullet\bullet}$.
Reference


*

*Misner, Thorne, Wheeler (1973): Gravitation (Freeman, reprint).

