Since the problem has a local nature, I'll consider only the Linear Algebra couterpart. Let $V$ be a real vector space whose dimension is $2n$ and let $$J\colon V \longrightarrow V$$ be a complex structure on $V$, $J^2 = -Id$. Then $(V,J)$ is a (n-dimensional) complex vector space for the scalar multiplication $$(a+bi)v = av+bJ(v).$$ Let $\left<,\right>$ be a (real) inner product on $V$ compatible with $J$, $$\left<J(v),J(w)\right> = \left<v,w\right>.$$ We define an hermitian product on $(V,J)$: $$\left(v,w\right) := \left<v,w\right> -i\left<J(v),w\right>.$$ On the other hand, we have a natural hermitian extension for $\left<,\right>$ to $V\otimes\mathbb{C}$: $$\left<v\otimes a,w \otimes b\right>_{\mathbb{C}} := a\bar b \left<v,w\right>.$$
Extending $J$ to $V\otimes\mathbb{C}$ we may take the eingenspaces for $J$ and make the decomposition $$V\otimes\mathbb{C} = V'\oplus V''.$$ Fix a (real) $\mathbb{C}-$basis $\{x_1, \dots x_n\}$ for $(V,J)$. Then the $y_j := J( x_j)$ complete a real basis for $V$. Define $$z_j = \frac{1}{2}(x_j -iy_j).$$ Since $J(z_j) =iz_j$, the set $\{z_1, \dots z_n\}$ is a $\mathbb{C}-$basis for $(V',i)$, the set $\{\bar z_1, \dots \bar z_n\}$ is a $\mathbb{C}-$basis for $(V'',-i)$ and $x_j \mapsto z_j$ , $x_j \mapsto \bar z_j$ are isomorphisms. Now we head to the product.
$$ \left<z_j,z_k\right>_{\mathbb{C}} = \frac{1}{4}\left[\left<x_j,x_k\right> -i\left<y_j,x_k\right> +i \left<x_j,y_k\right> +\left<y_j,y_k\right> \right] = \frac{1}{2}\left(x_j, x_k\right),
$$
$$ \left<z_j,\bar z_k\right>_{\mathbb{C}} = \frac{1}{4}\left[\left<x_j,x_k\right> -i\left<y_j,x_k\right> -i \left<x_j,y_k\right> -\left<y_j,y_k\right> \right] = 0,
$$
$$ \left<\bar z_j,\bar z_k\right>_{\mathbb{C}} = \overline{\left<z_j,z_k\right>}_{\mathbb{C}}.
$$
We have then the relation between the hermitian products $\left(,\right)$ on $(V,J)$ and $\left<,\right>_{\mathbb{C}}$ on $(V\otimes \mathbb{C},i)$. The last thing to remark is that on Kobayashi's notation $$ h(z_j,z_k) = \left<z_j,\bar z_k\right>_{\mathbb{C}}.$$
Best Answer
Yes it does. More generally, let $\overline{\mathbb{C}}^n$ be ${\mathbb{C}}^n$ with the conjugate complex structure, then the standard hermitian metric on $\mathbb C^n$ is an element $h_0$ of the vector space of sesquilinear forms $S(n) = (\mathbb C^n)^*\otimes_{\mathbb C} (\overline{\mathbb{C}}^n)^*$.
$U(n)$ acts on $S(n)$ by $A\cdot s(X,Y) = s(A^{-1}X, A^{-1}Y)$ and, most importantly, fixes $h_0$.
Consider now a principal $U(n)$ bundle $P$, then we have the associated vector bundles $E = P\times_{U(n)} \mathbb C^n$ and $T= P\times_{U(n)}S(n)$.
Since $h_0$ is fixed by $U(n)$, it defines a section of $T$, this is defined by $h_0$ in a local frame of $T$ induced by a local section of $P$, and the definition globalizes because $U(n)$ fixes $h_0$ (just check how things transform using the transformation functions of $T$ induced by $P$, see also this question Associated bundles). $h$ gives an hermitian metric on $E$ because $T$ is naturally isomorphic to the bundle of sesquilinear forms of $E$.
Moreover, $h$ will also be parallel with respect to any connection on $T$ induced by a connection on $P$, see for example the question Tensor fields defining $G$-structure are parallel.