For an $2n$-dimensional complex manifold $M$, and a smooth vector bundle $E$ over $M$, it is well-known (see Voisin, Huybrechts) that there exists an operator $\overline{\partial}$, built locally from the usual anti-holomorphic derivative, that acts on $\Gamma^{\infty}(E) \otimes\_{C^{\infty}} \Omega^{(0,\bullet)}(M)$ so as to give a complex
$$
\Gamma^{\infty}(E) \overset{\overline{\partial}}{\to} \Gamma^{\infty}(E) \otimes\_{C^{\infty}} \Omega^{(0,1)}(M) \overset{\overline{\partial}}{\to} \cdots \overset{\overline{\partial}}{\to} \Gamma^{\infty}(E) \otimes\_{C^{\infty}} \Omega^{(0,n)}(M)\overset{\overline{\partial}}{\to} 0.
$$


As I prefer global constructions, I began to wonder how one would construct this globally. To apply id $\otimes \overline{\partial}$ to $\Gamma^{\infty}(E) \otimes\_{C^{\infty}} \Omega^{(0,\bullet)}(M)$ is of course not well-defined since we are tensoring over ${C^{\infty}(M)}$. So what can one do . . .?


P.S. Does such a complex exist in the purely real case, and if not, then why not?



Global Definition of the Dolbeault Complex of a Vector Bundle
To get the Dolbeault complex, you need a choice of *holomorphic* structure on $E$, not just a smooth one. If $\mathcal{E}$ is the locally free sheaf of $\mathcal{O}\_M$-modules corresponding to $E$, then $\mathcal{E}\subset \mathcal{A}^0(\mathcal{E})=\mathcal{C}^\infty\_M\otimes\_{\mathcal{O}\_M}\mathcal{E}$ and
 $\overline{\partial}\_E: \mathcal{A}^0(\mathcal{E})\to \mathcal{A}^{0,1}(\mathcal{E})$
is the unique morphism (of sheaves of $\mathbb{C}$ vector spaces) satisfying
$$ \overline{\partial}\_E(f\sigma) = \overline{\partial}f\otimes \sigma + f\overline{\partial}\_E(\sigma),$$
for any smooth function $f$ and $\sigma$ a smooth section of $E$, such that 
$\left. \overline{\partial}\_E\right| \_{\mathcal{E}}=0$. The first term in the Leibniz formula involves $\overline{\partial}=d^{0,1}$.
You can then extend the Dolbeault operator to
$\overline{\partial}\_E: \mathcal{A}^{0,p}(\mathcal{E})\to \mathcal{A}^{0,p+1}(\mathcal{E})$,
$\overline{\partial}\_E^2=0$, 
by imposing the Leibniz rule with the usual sign. This gives you the Dolbeault resolution
$$ 0\to \mathcal{E}\to \mathcal{A}^0(\mathcal{E})\to \mathcal{A}^{0,1}(\mathcal{E})\to\ldots$$
The complex you write is obtained by passing to global sections of $\mathcal{A}^{0,\bullet}(\mathcal{E})$.


You cannot do any of this without the holomorphic structure. Differently put, you need
 the total space of $E$ to be a complex manifold and the projection $E\to M$ to be holomorphic. 
You cannot play the same game in the real case, but if you are willing to assume that $E$ carries a flat connection, then you can look at the de Rham resolution of the corresponding local system, as David explains.


**ADDENDUM**


The requirement that a (smooth) complex vector bundle $V$ admits a holomorphic structure
$\overline{\partial}\_E$ is non-trivial. It can be phrased as follows:


$V$ admits a holomorphic structure if and only if it admits a connection, $D$, such that
$D^{0,1}\circ D^{0,1}=0$, i.e., a connection for which the $(0,2)$ component of the curvature vanishes.


