= Solution
The <connection difference as an endomorphism-valued one-form> follows because the <graded Leibniz rule> terms cancel:
$$
(D_1-D)(fs)=f(D_1-D)s.
$$
Thus $a=D_1-D$ is a global endomorphism-valued one-form. To define the <endomorphism bundle connection>, require
$$
(\widetilde D\phi)(s)=D(\phi s)-\phi(Ds).
$$
The right side is $C^\infty$-linear in $s$, so it is indeed an endomorphism-valued one-form, and replacing $\phi$ by $f\phi$ gives $df\otimes\phi+f\widetilde D\phi$. In a frame with <connection matrix> $A$, its extension to an endomorphism-valued $k$-form $b$ is
$$
\widetilde D b=db+A\wedge b-(-1)^k b\wedge A.
$$
The <endomorphism-valued exterior product> combines the wedge of <differential forms> with composition in $\operatorname{End}E$. In particular,
$$
(a\wedge a)(u,v)=a(u)a(v)-a(v)a(u),\qquad
\widetilde D a=da+A\wedge a+a\wedge A.
$$
Expanding $d(A+a)+(A+a)\wedge(A+a)$ proves the <curvature difference formula>:
$$
\boxed{\Theta_{D+a}=\Theta_D+\widetilde D a+a\wedge a.}
$$
The <Chern connection> is characterized by <metric compatibility> and $D^{0,1}=\bar\partial_E$. Take <Hermitian inner products> linear in the first variable. In a <holomorphic local frame> write $\langle eu,ev\rangle=v^\dagger Hu$, with $H$ a positive <Hermitian matrix>. The second condition forces $A^{0,1}=0$, and <metric compatibility> is
$$
dH=A^\dagger H+HA.
$$
Its $(1,0)$ component determines $A$ uniquely:
$$
\boxed{A=H^{-1}\partial H.}
$$
Conversely this formula and its conjugate transpose satisfy the metric equation. Under a <holomorphic local frame> change, $H'=g^\dagger Hg$ and $\partial g^\dagger=0$, so the displayed formula transforms by $A'=g^{-1}Ag+g^{-1}\partial g$. Hence the local matrices patch, proving existence and uniqueness.
Differentiating $H^{-1}$ gives
$$
\partial A=-H^{-1}(\partial H)H^{-1}\wedge\partial H=-A\wedge A.
$$
Therefore the $(2,0)$ <vector-bundle curvature> is zero. There is no $(0,2)$ component, and
$$
\boxed{\Theta=\bar\partial A\in A^{1,1}(\operatorname{End}E),\qquad
\bar\partial_{\operatorname{End}E}\Theta=0.}
$$
The last equality is $\bar\partial^2 A=0$ in <holomorphic local frames>, so is intrinsic. This is <Dolbeault closedness of Chern curvature>. The induced <endomorphism bundle connection> is itself the <Chern connection> for the Hilbert-Schmidt metric on $\operatorname{End}E$: the <tensor product> of a metric-compatible connection and its dual preserves the induced metric, and its $(0,1)$ part is the induced holomorphic <Dolbeault operator>. These two properties and uniqueness suffice; no further matrix calculation is needed.
For two metrics, both <Chern connections> have the same $(0,1)$ part, so \b[$a\in A^{1,0}(\operatorname{End}E)$]. The $(1,1)$ part of the <curvature difference formula> is just $\bar\partial_{\operatorname{End}E}a$; the other two terms have type $(2,0)$, which must cancel because both <vector-bundle curvatures> have type $(1,1)$. Thus
$$
\boxed{\Theta_1-\Theta_0=\bar\partial_{\operatorname{End}E}a.}
$$
By <Dolbeault theorem>, this proves that the <curvature Dolbeault class>
$$
\alpha(E)=[\Theta]\in H^1(M,\Omega_M^1\otimes\operatorname{End}E)
$$
is independent of the metric.
Endomorphism composition and <wedge product of differential forms> define the powers in $A^{k,k}(\operatorname{End}E)$, rather than in a <tensor power> of the <endomorphism> bundle. The <Dolbeault operator> is a derivation, so $\bar\partial(\Theta^k)=0$. The noncommutative telescoping identity gives the stronger explicit independence statement
$$
\Theta_1^k-\Theta_0^k
=\sum_{j=0}^{k-1}\Theta_1^j\wedge(\Theta_1-\Theta_0)
\wedge\Theta_0^{k-1-j}
=\bar\partial_{\operatorname{End}E}
\left(\sum_{j=0}^{k-1}\Theta_1^j\wedge a\wedge\Theta_0^{k-1-j}\right).
$$
All powers of <vector-bundle curvature> have even total degree, so the derivation introduces no extra signs before $a$. Consequently
$$
\boxed{\alpha(E)^k=[\Theta^k]\in H^k(M,\Omega_M^k\otimes\operatorname{End}E)}
$$
is metric independent. For $k>n$, the form space and the resulting class are zero. This proves <metric independence of curvature Dolbeault powers> without assuming that endomorphism-valued forms commute.
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