An orthogonal or hermitian structure on $E$ is a section of *fibre bundle*
(which is not a vector bundle). I will deal with the complex case.

The Lie algebra $\mathfrak{gl}(n)$
decomposes into
$$
\mathfrak{gl}(n)\simeq \mathfrak{u}(n)\oplus \textrm{Herm}_n,
$$
where $\textrm{Herm}_n$ is the vector space of Hermitian $n\times n$ matrices. The exponential map sends
it to the positive-definite hermitian matrices:
$$
\exp: \textrm{Herm}_n\to \textrm{Herm}_n^+,
$$
which form a convex domain ("cone") . More importantly, this cone is actually
$$
\textrm{Herm}_n^+ = GL(n,\mathbb{C})/U(n).
$$
To see this, consider the action of $GL(n,\mathbb{C})$ on the hermitian matrices by
$(T,h)\mapsto \overline{T}^t h T$.
Now, do all of this "fibrewise": if $P$ is the frame bundle of $E$, an hermitian metric is a section of the associated bundle with fibre $GL(n,\mathbb{C})/U(n)$.

More coneptually, a choice of reduction of the structure group of a principal $G$-bundle $P$ to a subgroup $K$ is eqivalent to a choice of section of the associated $G/K$ bundle.
The hermitian metric is a reduction of the structure group from $GL(n,\mathbb{C})$ to $U(n)$.

**Aside:**

Since this question may be a related to your other question

Hermitian Christoffel Symbols

let
me say that if you have a complex manifold $X=(M,I)$ with Riemannian metric $g$ on $M$
(compatible with $I$) you can extend $g$ to the complexified tangent bundle $T_{X,\mathbb{C}}$ either as a $\mathbb{C}$-*bilinear* pairing (as in Kobayashi & Nomizu), or as a *sesquilinear*
pairing $g_{\mathbb{C}}$: see section 1.2 of Huybrechts, "Complex geometry". Now,
$T_{X,\mathbb{C}}\simeq T^{1,0}\oplus T^{0,1}$. With the former choice $T^{p,q}$
are isotropic sub-bundles, and only the off-diagonal pairing is non-trivial. With the latter choice (Huybrechts, Griffiths & Harris), the two subbundles are *orthogonal*
and
$\left. g_{\mathbb{C}}\right|_{T^{1,0}}$ is $\frac{1}{2}h$, $h= g-i\omega$ (and the conjugate of that on $T^{0,1}$). This turns $E = T^{1,0}$ into an hermitian vector bundle.

In indices, in the first case you have $h_{ab}=0= h_{\overline{a}\overline{b}}$,
$h_{a\overline{b}}\neq 0$, while in the second the other way around.

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