= Solution
The <Kähler form> and the <Hermitian metric> give the $L^2$ inner product on <vector-bundle-valued differential forms>, using volume $\omega^n/n!$. Define the <formal adjoint> $\bar\partial_E^*$ and the elliptic, self-adjoint, nonnegative <Dolbeault Laplacian>
$$
\Delta''_E=\bar\partial_E\bar\partial_E^*+\bar\partial_E^*\bar\partial_E.
$$
Its harmonic space is
$$
\mathcal H^{p,q}(X,E)=\ker\Delta''_E
=\ker\bar\partial_E\cap\ker\bar\partial_E^*.
$$
The equality follows from $\langle\Delta''_E\alpha,\alpha\rangle=\|\bar\partial_E\alpha\|^2+\|\bar\partial_E^*\alpha\|^2$. On compact $X$, the bundle-valued <Dolbeault Hodge decomposition> states that this space is finite dimensional and that
$$
\boxed{\mathcal A^{p,q}(X,E)=\mathcal H^{p,q}(X,E)
\oplus\bar\partial_E\mathcal A^{p,q-1}(X,E)
\oplus\bar\partial_E^*\mathcal A^{p,q+1}(X,E).}
$$
The sum is orthogonal for the $L^2$ inner product and all summands here consist of smooth forms. Every <Dolbeault cohomology> class has a unique harmonic representative, giving $H^{p,q}_{\bar\partial}(X,E)\cong\mathcal H^{p,q}(X,E)$ and, by the <Dolbeault theorem>, the corresponding <sheaf cohomology> isomorphism. This is a decomposition for $\bar\partial_E$, whose square is zero; it does not require the full <Chern connection> to be flat.
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