A theorem of Demailly and Pǎun (https://arxiv.org/abs/math/0105176) gives a condition for a compact complex manifold to be of Fujiki class $\mathcal{C}$ (that is bimeromoprhic to a Kähler manifold). A manifold is Kähler if and only if it supports a Kähler current. A Kähler current is a real positive $(1,1)$-current $T$, such that $T - \varepsilon \omega > 0$ for a constant $\varepsilon > 0$ and positive hermitian $(1,1)$-form $\omega$.
The homology class of such current is integral if and only if the manifold is projective (this is the standard Kodaira embedding theorem).
A similar theorem for Moishezon manifolds was proved by Popovici (https://arxiv.org/abs/math/0603738). It states, roughly speaking, that a complex manifold is Moishezon if and only if it supports a semi-positive and strictly positive almost everywhere real $(1,1)$-current with integral homology class.
The integrality condition is hardly a differential-geometric one, but you cannot escape it (at least, as far, is we know by now), and I don't really think one might give a more differential-geometric answer.
In a general situation it is, of course, rather hard to verify the Popovici condition. Another "geometric" approach to studying algebraic dimensions of complex manifolds is to study the algebraic reduction. For any complex manifold $X$ there exists a normal projective variety $\overline{X}$ and a meromorphic map $\alpha \colon X \to \overline{X}$, such that any meromorphic function on $X$ can be lifted from $\overline{X}$. The variety $\overline{X}$ is unique up to birational equivalence.
Being Moishezon is equivalent to $\alpha$ being a birational equivalence. More generally, $a(X) = \dim_{\mathbb C}(\overline{X})$.
This is correct. However, when people say "compact manifold" they almost always mean connected compact manifold. Rather, there is usually nothing to be gained by dealing with non-connected compact manifolds, since we might as well just look at each connected component.
(For non-compact manifolds, this is potentially trickier, because we have things like $(-\infty,0)\cup(0,\infty)$ which is a disjoint union of two manifolds, but they are sort of "touching," and in some sense inherently different from $(-\infty,-1)\cup(1,\infty)$, for example.)
Best Answer
Consider $\int_X d\eta\wedge\overline{d\eta}$. Check that the integrand is exact, since $d\eta = \partial\eta$ and $d\overline{d\eta} = d\bar\partial\bar\eta = \partial\bar\partial\bar\eta = -\bar\partial\partial\bar\eta = -\bar\partial\overline{\bar\partial\eta} = 0$.