Let me try to clarify the point that is bothering you. I think it's best to look at it in the simplest perspective: linear algebra, because it all boils down to linear algebra.
Here's our situation: we have a real vector space $V$ with a linear complex structure $J : V \to V$.
Definition 1. An inner product $g$ on $V$ is said to be compatible with the linear complex structure $J$ if $J$ is orthogonal with respect to $g$ (i.e. $g(Jx,Jy) = g(x,y)$ for all $x,y \in V)$.
Note that this is what you call a Hermitian inner product but I prefer to avoid this terminology for the sake of clarity. Instead:
Definition 2. A Hermitian inner product on $(V,J)$ is a real bilinear map $h : V \times V \to \mathbb{C}$ which is sesquilinear in the sense that $h(Jx, y) = -h(x, Jy) = ih(x,y)$ for all $x,y \in V$ and Hermitian-symmetric in the sense that $h(y,x) = \overline{h(x,y)}$ for all $x, y \in V$.
Now, the point is that these two notions are essentially the same:
Proposition. If $h$ is a Hermitian inner product on $(V,J)$, then $g := \operatorname{Re}(h)$ is a compatible inner product on $(V,J)$. Conversely, is $g$ is a compatible inner product on $(V,J)$, then there is a unique Hermitian inner product $h$ on $(V,J)$ such that $g = \operatorname{Re}(h)$.
As you can see, this has absolutely nothing to do with the complexification of $V$. It's true that $g$ extends to a symmetric complex bilinear map $g^\mathbb{C} : V^\mathbb{C} \to V^\mathbb{C}$ where $V^\mathbb{C} = V \otimes_{\mathbb{R}} \mathbb{C}$, but it's a different subject (and has nothing to do with the complex structure $J$).
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
K., this stuff confuses me too, but here's how I understand it. Consider $X$ as a real, $2n$ dimensional manifold equipped with a (Riemannian) metric $g$, and a complex stucture $J$. You are correct in saying it is called Hermitian if $g(JX,JY) = g(X,Y)$, but this does not make it a hermitian inner product in the usual sense.
Now for any $p\in X$ $J_p: T_pX \rightarrow T_pX$ satisfies $J_p^2=-Id$, and so we may always choose real coordinates $x_1,y_1,\ldots,x_n,y_n$ on $X$ such that $J(\frac{\partial}{\partial x}) = \frac{\partial}{\partial y}$ and $J(\frac{\partial}{\partial y}) = -\frac{\partial}{\partial x}$. So here's how you get $J$ in local coordinates.
If we extend $J_p$ to the complexification of $T_pX$ it will have two eigenvalues $i$ and $-i$ and $T_pX\otimes\mathbb{C}$ will split into two eigenspaces: $T_pX\otimes\mathbb{C}=T^{'}_pX\oplus T^{''}_pX$. The $i$ eigenspace, $T^{'}_pX$, is the holomorphic tangent space and is spanned by vectors $\frac{\partial}{\partial z_i} = \frac{\partial}{\partial x_i}-i\frac{\partial}{\partial y_i}$ while $T^{''}_pX$ is spanned by $\frac{\partial}{\partial\bar{z}_i} = \frac{\partial}{\partial x_i}+i\frac{\partial}{\partial y_i}$. Note that if $\xi \in T_p^{'}X$ $\bar{\xi}\in T_p^{''}$.
We can extend $g$ by complex linearity to be defined on $T_pX\otimes\mathbb{C}$. In coordinates dual to the basis $\{\frac{\partial}{\partial z_i}, \frac{\partial}{\partial \bar{z_j}}\}$ we can write this as:
$$g = \sum g_{ij}dz_i\otimes dz_j + \sum g_{\bar{i}\bar{j}}\bar{dz}_i\otimes\bar{dz}_j + \sum g_{\bar{i}j} \bar{dz_i}\otimes dz_j + \sum g_{j\bar{i}} dz_j\otimes\bar{dz_i}$$
As in Javier's comment, where $g_{ij} = g(\frac{\partial}{\partial z_i},\frac{\partial}{\partial z_j})$ and so on.
Observe that
$$ g(J\frac{\partial}{\partial z_i},J\frac{\partial}{\partial z_j}) = g(i\frac{\partial}{\partial z_i},i\frac{\partial}{\partial z_j}) = -g(\frac{\partial}{\partial z_i},\frac{\partial}{\partial z_j})= -g_{ij}$$
Using the fact that $g$ is Hermitian however we also have $$g(J\frac{\partial}{\partial z_i},J\frac{\partial}{\partial z_j})= g(\frac{\partial}{\partial z_i},\frac{\partial}{\partial z_j}) = g_{ij}$$
Hence $g_{ij} = 0$. Similarly $g_{\bar{i}\bar{j}} = 0$. Finally observe that:
$$ g_{\bar{i}j} = g(\frac{\partial}{\partial \bar{z_i}}, \frac{\partial}{\partial z_j}) = g(\frac{\partial}{\partial x_i},\frac{\partial}{\partial x_j}) + g(\frac{\partial}{\partial y_i}, \frac{\partial}{\partial y_j}) + i\left(g(\frac{\partial}{\partial y_i},\frac{\partial}{\partial x_j}) - g(\frac{\partial}{\partial x_i},\frac{\partial}{\partial y_j})\right) $$ But, using the fact that $g$ is Hermitian again, as well as the definition of $\frac{\partial}{\partial y_i}$ and $\frac{\partial}{\partial x_i}$: $$ g(\frac{\partial}{\partial y_i},\frac{\partial}{\partial x_j}) = g(J\frac{\partial}{\partial x_i},-J\frac{\partial}{\partial y_j}) = - g(\frac{\partial}{\partial x_i},\frac{\partial}{\partial y_j}) $$ We also have $$ g(\frac{\partial}{\partial x_i},\frac{\partial}{\partial x_j}) = g(\frac{\partial}{\partial y_i},\frac{\partial}{\partial y_j}) $$
and so:
$$g_{\bar{i}j} = 2g(\frac{\partial}{\partial x_i},\frac{\partial}{\partial x_j}) -2ig(\frac{\partial}{\partial x_i},\frac{\partial}{\partial y_j})$$
Using the same argument (i.e. using the fact that $g$ is Hermitian as well as the relationship between $\frac{\partial}{\partial x_i}$ and $\frac{\partial}{\partial y_i}$) we can show that: $$g_{j\bar{i}} = 2g(\frac{\partial}{\partial x_i},\frac{\partial}{\partial x_j}) -2ig(\frac{\partial}{\partial x_i},\frac{\partial}{\partial y_j}) = g_{\bar{i}j}$$ We can also show that
$$g_{i\bar{j}} = \overline{g_{\bar{i}j}} $$
So we have $g = 2\sum g_{\bar{i}j} (\bar{dz_i}\otimes dz_j+dz_j\otimes\bar{dz_i})$ as required.
A word of caution; $g$ as defined above is a Hermitian metric in the sense of Griffiths and Harris. That is, a "positive definite Hermitian inner product: $$ (\ ,\ )_{p}: T^{'}_{p}M\otimes\overline{T^{'}_{p}M} \to \mathbb{C}$$ on the holomorphic tangent space at $p$ for each $p\in M$ depending smoothly on $p$" (pg. 27). I find it clearer to think of
$$h_p: T^{'}_{p}M\otimes T^{'}_{p}M \to \mathbb{C}$$ $$h_p(\xi,\zeta) = g(\xi,\overline{\zeta})$$ as the Hermitian metric on $T^{'}_{p}M$ as this more obviously (to me at least) a Hermitian inner product on each tangent space.
I tried to find a good reference for this construction; I think the best would be either Huybrecht's Complex Geometry pg. 28-31 or Gross, Huybechts and Joyce's "Calabi-Yau Manifolds and Related Geometries" (look at the beginning of the chapter written by Joyce).