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).
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
The answer to your first question is yes -- regardless of whether the metric $g$ is compatible with $J$, the section $H \in \Gamma(T^*X^{\otimes 2} \otimes \mathbb{C})$ defined by $H(\lambda v, \mu w):=\tfrac{1}{2}\lambda\overline{\mu}g(v,w)$ is a Hermitian metric. This is just because it is a smoothly varying Hermitian inner product on each fiber of $TX \otimes \mathbb{C}$. That having been said, $H$ will not necessarily play nicely with the holomorphic and antiholomorphic bundles, commonly denoted by $T^{1,0}X$ and $T^{0,1}X$. In fact, these two subbundles are $H$-orthogonal to one another if and only if $g$ is $J$-compatible. In the case that $g$ is not $J$-compatible, we will also no longer have the nice decomposition $H= g+i\omega$, which is the standard shorthand for the situation (indeed, there is no reason for the form define by $$(v+iJv,w+iJw) \mapsto g(v,w) + ig(Jv,w)$$ to be Hermitian if $g$ is not assumed to be compatible with $J$). We will, however, still have an analogous decomposition $H= \tilde{g} + i\tilde{\omega}$, where $$\tilde{g}(v,w):=\tfrac{1}{2}\big(g(v,w)+g(Jv,Jw)\big)$$ and $$\tilde{\omega}(v,w):=\tfrac{1}{2}\big(g(Jv,w)-g(v,Jw)\big).$$ As before, we have that $g=\tilde{g}$ and $\omega=\tilde{\omega}$ iff $g$ is compatible with $J$.
Please let me know if I can help clarify any of this. Differential geometry is rife with abuses of notation. Of course, as soon as one understands what is meant by the abusive notation, one happily adopts it. I am no exception =).