[proofplan]
We define $\bar\partial_E$ locally by applying the scalar Dolbeault operator to the coefficient forms in a holomorphic frame. The only point requiring verification is that this coefficientwise prescription is unchanged when the holomorphic frame is replaced by another one. On an overlap, the two frames differ by a holomorphic transition matrix, and the scalar identity $\bar\partial g_{ij}=0$ removes the extra derivative terms. Once the local definitions agree on overlaps, they glue to a global operator, and the graded Leibniz rule follows coefficientwise from the scalar graded Leibniz rule for $\bar\partial$.
[/proofplan]
[step:Define the coefficientwise Dolbeault operator in one holomorphic frame]
Let $U\subset X$ be an [open set](/page/Open%20Set) on which $E$ admits a holomorphic frame $e=(e_1,\dots,e_m)$. Thus each $e_i:U\to E|_U$ is a holomorphic local section, and $(e_1(x),\dots,e_m(x))$ is a complex basis of the fiber $E_x$ for every $x\in U$.
Let $\alpha\in A^{p,q}(U,E)$. Because $e$ is a frame, there are unique scalar forms $\alpha_i\in A^{p,q}(U)$ such that
\begin{align*}
\alpha=\sum_{i=1}^{m}\alpha_i\otimes e_i.
\end{align*}
Define the local operator associated to the frame $e$ by the map
\begin{align*}
\bar\partial_{E,e}:A^{p,q}(U,E)\to A^{p,q+1}(U,E)
\end{align*}
given by
\begin{align*}
\bar\partial_{E,e}\alpha=\sum_{i=1}^{m}\bar\partial\alpha_i\otimes e_i.
\end{align*}
This is a well-defined $\mathbb{C}$-[linear map](/page/Linear%20Map) on $U$ because the coefficient forms $\alpha_i$ are uniquely determined by the frame $e$, and the scalar Dolbeault operator sends $A^{p,q}(U)$ to $A^{p,q+1}(U)$.
[/step]
[step:Compare the coefficientwise formulas on overlaps of holomorphic frames]
Let $V\subset X$ be another open set with holomorphic frame $e'=(e'_1,\dots,e'_m)$, and put $W:=U\cap V$. On $W$, define the transition matrix
\begin{align*}
G:W\to GL_m(\mathbb{C})
\end{align*}
by the equations
\begin{align*}
e'_j=\sum_{i=1}^{m}g_{ij}e_i
\end{align*}
where each entry $g_{ij}:W\to\mathbb{C}$ is holomorphic.
Write the same bundle-valued form on $W$ in the two frames:
\begin{align*}
\alpha|_W=\sum_{i=1}^{m}\alpha_i\otimes e_i=\sum_{j=1}^{m}\alpha'_j\otimes e'_j.
\end{align*}
Substituting the expression for $e'_j$ in the frame $e$ gives
\begin{align*}
\alpha_i=\sum_{j=1}^{m}g_{ij}\alpha'_j
\end{align*}
for each $i\in\{1,\dots,m\}$. Applying the scalar Dolbeault operator to this coefficient identity and using the scalar Leibniz rule with $g_{ij}$ of total degree $0$ gives
\begin{align*}
\bar\partial\alpha_i=\sum_{j=1}^{m}\bar\partial(g_{ij}\alpha'_j)=\sum_{j=1}^{m}\bigl(\bar\partial g_{ij}\wedge\alpha'_j+g_{ij}\bar\partial\alpha'_j\bigr).
\end{align*}
Since each $g_{ij}$ is holomorphic, $\bar\partial g_{ij}=0$. Hence
\begin{align*}
\bar\partial\alpha_i=\sum_{j=1}^{m}g_{ij}\bar\partial\alpha'_j.
\end{align*}
Therefore
\begin{align*}
\sum_{i=1}^{m}\bar\partial\alpha_i\otimes e_i=\sum_{i=1}^{m}\sum_{j=1}^{m}g_{ij}\bar\partial\alpha'_j\otimes e_i=\sum_{j=1}^{m}\bar\partial\alpha'_j\otimes e'_j.
\end{align*}
Thus $\bar\partial_{E,e}\alpha=\bar\partial_{E,e'}\alpha$ on $W$.
[guided]
The possible obstruction to well-definedness is the following: the coefficients of $\alpha$ change when the frame changes, so applying $\bar\partial$ to those coefficients might produce extra terms. We must show that holomorphicity of the transition matrix makes those extra terms vanish.
Let $W:=U\cap V$. The two holomorphic frames $e=(e_1,\dots,e_m)$ and $e'=(e'_1,\dots,e'_m)$ are related by a transition matrix
\begin{align*}
G:W\to GL_m(\mathbb{C}).
\end{align*}
By definition of the entries of $G$, there are holomorphic functions $g_{ij}:W\to\mathbb{C}$ such that
\begin{align*}
e'_j=\sum_{i=1}^{m}g_{ij}e_i
\end{align*}
for each $j\in\{1,\dots,m\}$.
Now write the same form in both frames:
\begin{align*}
\alpha|_W=\sum_{i=1}^{m}\alpha_i\otimes e_i=\sum_{j=1}^{m}\alpha'_j\otimes e'_j.
\end{align*}
Substituting the expression for $e'_j$ into the second expansion expresses the primed coefficients in the unprimed frame:
\begin{align*}
\sum_{j=1}^{m}\alpha'_j\otimes e'_j=\sum_{j=1}^{m}\alpha'_j\otimes\sum_{i=1}^{m}g_{ij}e_i.
\end{align*}
Collecting the coefficient of $e_i$ and using uniqueness of coefficients in the frame $e$, we obtain
\begin{align*}
\alpha_i=\sum_{j=1}^{m}g_{ij}\alpha'_j.
\end{align*}
We now apply the scalar Dolbeault operator to this identity. The scalar graded Leibniz rule applies to the product of the function $g_{ij}$, which has total degree $0$, and the scalar form $\alpha'_j\in A^{p,q}(W)$. Thus
\begin{align*}
\bar\partial(g_{ij}\alpha'_j)=\bar\partial g_{ij}\wedge\alpha'_j+g_{ij}\bar\partial\alpha'_j.
\end{align*}
The reason the frame must be holomorphic is exactly here: because $g_{ij}$ is holomorphic, its antiholomorphic derivative vanishes:
\begin{align*}
\bar\partial g_{ij}=0.
\end{align*}
Therefore
\begin{align*}
\bar\partial(g_{ij}\alpha'_j)=g_{ij}\bar\partial\alpha'_j.
\end{align*}
Summing over $j$ gives
\begin{align*}
\bar\partial\alpha_i=\sum_{j=1}^{m}g_{ij}\bar\partial\alpha'_j.
\end{align*}
Finally we compare the two local formulas:
\begin{align*}
\bar\partial_{E,e}\alpha=\sum_{i=1}^{m}\bar\partial\alpha_i\otimes e_i.
\end{align*}
Using the just-proved coefficient transformation law,
\begin{align*}
\bar\partial_{E,e}\alpha=\sum_{i=1}^{m}\sum_{j=1}^{m}g_{ij}\bar\partial\alpha'_j\otimes e_i.
\end{align*}
Reordering the finite sum and using $e'_j=\sum_{i=1}^{m}g_{ij}e_i$, this becomes
\begin{align*}
\bar\partial_{E,e}\alpha=\sum_{j=1}^{m}\bar\partial\alpha'_j\otimes e'_j.
\end{align*}
The right-hand side is precisely $\bar\partial_{E,e'}\alpha$. Hence the two framewise definitions agree on $W$.
[/guided]
[/step]
[step:Glue the local formulas to a global operator]
The preceding overlap computation shows that for any two holomorphic frames $e$ and $e'$ over open sets $U$ and $V$, the local forms $\bar\partial_{E,e}(\alpha|_U)$ and $\bar\partial_{E,e'}(\alpha|_V)$ agree on $U\cap V$. Since holomorphic frames exist locally for a holomorphic vector bundle, these compatible local forms glue uniquely to a global element of $A^{p,q+1}(X,E)$. Define this glued form to be $\bar\partial_E\alpha$.
The construction is local and coefficientwise $\mathbb{C}$-linear, so $\bar\partial_E:A^{p,q}(X,E)\to A^{p,q+1}(X,E)$ is a well-defined $\mathbb{C}$-linear operator.
[/step]
[step:Prove the graded Leibniz rule coefficientwise in a holomorphic frame]
Let $\beta\in A^{p,q}(X)$ and let $\alpha\in A^{r,s}(X,E)$. It is enough to verify the identity on an arbitrary open set $U\subset X$ admitting a holomorphic frame $e=(e_1,\dots,e_m)$. Write
\begin{align*}
\alpha|_U=\sum_{i=1}^{m}\alpha_i\otimes e_i
\end{align*}
with $\alpha_i\in A^{r,s}(U)$. Then
\begin{align*}
(\beta\wedge\alpha)|_U=\sum_{i=1}^{m}(\beta|_U\wedge\alpha_i)\otimes e_i.
\end{align*}
Applying the definition of $\bar\partial_E$ in the frame $e$ gives
\begin{align*}
\bar\partial_E(\beta\wedge\alpha)|_U=\sum_{i=1}^{m}\bar\partial(\beta|_U\wedge\alpha_i)\otimes e_i.
\end{align*}
The scalar graded Leibniz rule for the Dolbeault operator, applied to $\beta|_U\in A^{p,q}(U)$ and $\alpha_i\in A^{r,s}(U)$, gives
\begin{align*}
\bar\partial(\beta|_U\wedge\alpha_i)=\bar\partial(\beta|_U)\wedge\alpha_i+(-1)^{p+q}\beta|_U\wedge\bar\partial\alpha_i.
\end{align*}
Substituting this into the previous expression,
\begin{align*}
\bar\partial_E(\beta\wedge\alpha)|_U=\sum_{i=1}^{m}\bigl(\bar\partial(\beta|_U)\wedge\alpha_i\bigr)\otimes e_i+(-1)^{p+q}\sum_{i=1}^{m}\bigl(\beta|_U\wedge\bar\partial\alpha_i\bigr)\otimes e_i.
\end{align*}
The first sum is $(\bar\partial\beta\wedge\alpha)|_U$, and the second sum is $(\beta\wedge\bar\partial_E\alpha)|_U$. Hence
\begin{align*}
\bar\partial_E(\beta\wedge\alpha)|_U=(\bar\partial\beta\wedge\alpha)|_U+(-1)^{p+q}(\beta\wedge\bar\partial_E\alpha)|_U.
\end{align*}
Since $U$ was arbitrary among holomorphic frame domains and these domains cover $X$, the desired global identity follows:
\begin{align*}
\bar\partial_E(\beta\wedge\alpha)=\bar\partial\beta\wedge\alpha+(-1)^{p+q}\beta\wedge\bar\partial_E\alpha.
\end{align*}
[/step]
