Given a complex vector space $V$ and a Hermitian form $h$ on it, when people write $h = g + i \omega$ saying that $g$ is the real part of $h$ and $\omega$ the imaginary part, what are exactly $g$ and $\omega$ (i.e. how are they defined and on what space)?
The obvious answer is $g(u, v) = \Re(h(u,v))$ and $\omega=\Im(h(u,v))$. This makes the equality $h = g + i \omega$ obvious.
Why is this answer problematic? In Kahler geometry, people write $h = g + i \omega$ and specify that $\omega$ is the $(1, 1)$-Kahler form and often say nothing about $g$ (see this Wikipedia article for an example). This makes no sense to me because $\omega$ is a $\mathbb{C}$-bilinear form and $\Im(h)$ is not. What to me would make sense to say, is that on a Hermitian manifold $h = \Re h + i \Im h$ is such that $\Re h$ restricts to a Riemannian metric $g$ on the underlying real manifold and $\Im h$ restrict to an almost symplectic form, which in turn extends to a $(1,1)$-form, but $\Im h$ and $\omega$, that are both defined on the complexified tangent bundle, are not the same thing.
In case what I am saying is right I do not understand the sentence "The Hermitian metric $h$ can be recovered from $g$ and $ω$ via the identity $h = g + i \omega$" from the linked Wikipedia article.
Consider a complex vector space $V$ equipped with a Hermitian inner product $h$, that we take to be conjugate-linear in the second component. $V$ has an underlying real vector space $V_{\mathbb{R}}$, which is just the same set with the same operations, but which we restrict to consider only multiplication by real scalars.
The space $V_{\mathbb{R}}$ inherits a few structures from $(V,h)$; we have
To check that $g$ is symmetric and $\omega$ is skew symmetric is a quick exercise; let's do it for $\omega$. $$\omega(v,w)=\Im(h(v,w))=\Im(\overline{h(w,v)})=-\Im(h(w,v))=-\omega(w,v).$$ These two objects are real, defined on the real vector space $V_{\mathbb{R}}$, and as they stand are just $\mathbb{R}$-linear. Indeed, complex multiplication in $V_{\mathbb{R}}$ (which is given by $J$) exchanges $g$ and $\omega$, so surely it cannot preserve them. This is the way you should interpret the formula $h=g+\sqrt{-1}\,\omega$.
What often happens in Kähler geometry is that you usually extend $g$ and $\omega$ to $V_{\mathbb{C}}:=V_{\mathbb{R}}\otimes\mathbb{C}$, which is however quite different from the original $V$: it has double the (complex) dimension!
To say that $\omega$ is of type $(1,1)$ is to say that its complex linear extension is, i.e. $\omega$ vanishes when evaluated on two elements of $V_{\mathbb{C}}$ that are both of type $(1,0)$ or both of type $(0,2)$. Notice however that this can be detected already on $V_{\mathbb{R}}$, as $\omega$ is of type $(1,1)$ if and only if for every $v,w\in V_{\mathbb{R}}$ one has $\omega(Jv,Jw)=\omega(v,w)$. We can check this by direct computation $$\omega(Jv,Jw)=\Im\left(h(\sqrt{-1}\,v,\sqrt{-1}\,w)\right)=\Im\left(h(v,w)\right)=\omega(v,w).$$