This is proposition 7.24 on Kobayashi's Differential geometry of complex vector bundles. I get stuck understanding one part of the proof, which I will reproduce below.
Prop: A closed real $(p,p)$-form $\omega$ on a compact Kähler manifold $M$ is cohomologous to zero if and only if $\omega = id'd''\phi$ for some real $(p-1,p-1)$-form $\phi$.
Proof: assume that $[\omega]=0$. Then it is exact $\omega = d\alpha$ for some $(2p-1)$-form $\alpha$ and because $\omega$ is real, we deduce that the imaginary part of $\alpha$ is closed and we can throw it away and assume that $\alpha$ itself is real. Because of the bidegree decomposition $d=d'+d''$, $\alpha$ must lie in $\Omega_\mathbb{C}^{p-1,p}\oplus \Omega_\mathbb{C}^{p,p-1}$, so $$ \alpha = \beta_1 \oplus \beta_2, $$ but since conjugation exchanges bidegrees we deduce by $\bar{\alpha} = \alpha$ that $\bar{\beta_1}=\beta_2$ and we can assume $$ \alpha = \beta + \bar{\beta} $$ for $\beta$ a $(p-1,p$-form. Imposing $\omega=d\alpha$ and using $d=d'+d''$ gives $$ \omega = d'\beta + d''\beta + d'\bar{\beta} + d''\bar{\beta} $$ but since $\omega$ is of bidegree $(p,p)$, the first and last term must vanish. This is where I get stuck: Kobayashi argues that since $d''\beta=0$ it must be written as $$\beta = H\beta + d''\gamma$$ where $H\beta$ is the harmonic part and $\gamma$ is a $(p-1,p-1)$-form. This $\gamma$ will be then used to construct the $\phi$ in the statement of the theorem, however, Kobayashi doesn't clarify why such a decomposition for $\beta$ exist nor which Laplacian is he using to compute the "harmonic part". I have not yet reached the study of the Hodge decomposition, so at this point I don't know how to understand this decomposition.
Could it be achieved by means of some $\partial$, $\bar{\partial}$- Poincaré-type lemma or is this statement inherent to a Harmonic point of view?
Notation: $d',d''$ correspond to $\partial,\bar{\partial}$ in some other text-books, i.e. the (+1,0) and $(0,+1)$ projection of the classical exterior derivative.