= Solution
The <orthogonal projection> $\pi_S$ is a smooth bundle map, since the <Hermitian metric> and the subbundle vary smoothly. The map
$$
E/S\longrightarrow S^\perp,\qquad [v]\longmapsto(I-\pi_S)v
$$
is independent of the representative, bijective on each fibre, and smooth with smooth inverse supplied by the quotient map on $S^\perp$. Hence it is a natural isomorphism of smooth <vector bundles>. It need not be holomorphic. The <quotient Hermitian metric> is
$$
\boxed{h_Q([v],[w])=h_E((I-\pi_S)v,(I-\pi_S)w).}
$$
It is well-defined and positive definite because $S^\perp$ represents each quotient class uniquely.
On sections of $S$, set $D'=\pi_S D_E$. The <Leibniz rule> makes $D'$ a <connection on a vector bundle>. For $s,t\in\Gamma(S)$, the orthogonal terms vanish in the metric pairing, so
$$
dh_S(s,t)=h_S(D's,t)+h_S(s,D't).
$$
Thus $D'$ has <metric compatibility>. Because $S$ is a holomorphic subbundle, the $(0,1)$ part of $D_E$ preserves $S$ and restricts to $\bar\partial_S$. Therefore $(D')^{0,1}=\bar\partial_S$. The uniqueness of the <Chern connection> in part (b) gives the <projected Chern connection> formula
$$
\boxed{D_S=\pi_S D_E.}
$$
It follows that $A(s)=(I-\pi_S)D_Es$ is a quotient-valued one-form. \b[The printed target $T_M\otimes Q$ is missing a dual:] the <cotangent bundle> is the correct factor,
$$
\boxed{A(s)\in\Gamma(T_M^*\otimes Q).}
$$
Complex-valued forms use the complexified cotangent bundle here. In fact the $(0,1)$ part cancels, so $A$ is the <second fundamental form of a holomorphic subbundle>, an element of $\mathcal A^{1,0}(\operatorname{Hom}(S,Q))$. Finally the two connection <Leibniz rules> give
$$
A(fs)=df\otimes s+fD_Es-df\otimes s-fD_Ss
=\boxed{fA(s)}.
$$
This proves its <tensoriality> and all the asserted quotient and projection properties.
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