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).
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 $J$).
Best Answer
This stuff is really confusing, so don't feel bad about it. I find it always helpful to think of real and complex vector spaces first, before passing to complex manifolds.
So let's start with a real vector space $V$, and suppose we have a complex structure $I \in \operatorname{End}(V)$, i.e. $I^2 = - \operatorname{id}_V$. In that way we may also consider $V$ as a complex vector space. Let $g: V \times V \to \mathbb R$ be a real inner product, i.e. a positive definite, symmetric real-bilinear form, and suppose that $g$ is hermitian in the sense of Huybrechts, i.e. $$g(v,w) = g(I(v), I(w)) \tag{I}$$ for all $v,w \in V$. Then we may define $$h(v,w) = g(v,w) + i g(v, I(w)).$$ Claim. The form $h$ is a Hermitian inner product on $V$, i.e. it is a positive definite, sesquilinear form $V \times V \to \mathbb C$.
Proof. First note that by (I), we have $g(v, I(v)) = g(I(v), I^2(v)) = - g(I(v), v) = - g(v, I(v))$, so that $g(v, I(v)) = 0$. In particular $$h(v,v) = g(v,v) \geq 0$$ with equality if and only if $v = 0$. Hence $h$ is positive definite.
By a similar application of (I) and the symmetry of $g$ we have $$h(v, w) = g(v, w) + i g(v, I(w)) = g(w, v) - ig(w, I(v)) = \overline{h(w, v)}.$$ So it remains to show that $h$ is $\mathbb C$-linear in the first component. It is clearly additive, since $g$ is additive in the first component. Now if $\lambda = \mu + i \nu \in \mathbb C$ is a scalar, we get \begin{align*} h(\lambda v, w) & = g(\lambda v, w) + i g(\lambda v, I(w)) \\ & = \mu g(v, w) + \nu g(I(v), w) + i \mu g(v, I(w)) + i \nu g(I(v), I(w)) \\ & \stackrel{\text{(I)}}{=} \left(\mu g(v, w) + i \nu g(v, w) \right) + i\left( \mu g(v, I(w)) + i \nu g(v, I(w)) \right)\\ & = \lambda g(v, w) + i \lambda g(v, I(w)) \\ & = \lambda h(v, w). \end{align*}
Now the same applies to complex manifolds, where you replace $V$ with the real tangent space $T_x X$, and the real inner product with the Riemannian metric. So we get that every Riemannian metric, that is hermitian in the sense of (I), defines a Hermitian inner product on all tangent spaces. This is called a hermitian form, I think. Conversely, given a hermitian inner product $h$, you can always obtain $g$ as the real part of $h$, and by sesquilinearity, (I) will hold for such $g$.