= Solution
The specified bidegree is the <conjugate-linear Hodge star>, not the complex-linear extension of the real star. If the pointwise <Hermitian inner product> is linear in its first argument, it is defined by
$$
\alpha\wedge *\beta=\langle\alpha,\beta\rangle\,dV_g,\qquad *:A^{p,q}\longrightarrow A^{n-p,n-q}.
$$
It is the real <Hodge star operator> extended complex-linearly and composed with complex conjugation; $*^2=(-1)^{p+q}$. On a compact <Hermitian manifold>,
$$
\boxed{(\alpha,\beta)_{L^2}=\int_M\alpha\wedge *\beta}
$$
is a positive definite <Hermitian inner product> on smooth forms. The formal adjoint is $\bar\partial^*=-*\bar\partial*$ in this convention; the final star in this identity is present in the PDF but is corrupted in the TeX aid.
State the relevant <Hodge theorem> as follows. On a compact manifold without boundary, the <Dolbeault Laplacian> $\Delta_{\bar\partial}=\bar\partial\bar\partial^*+\bar\partial^*\bar\partial$ is elliptic, self-adjoint and nonnegative. Its harmonic kernel $\mathcal H^{p,q}_{\bar\partial}$ is finite-dimensional, and smooth forms split orthogonally as this kernel plus the image of $\Delta_{\bar\partial}$. There are the orthogonal projection $H$ and a smooth Green operator $G$ with
$$
\Delta_{\bar\partial}G=G\Delta_{\bar\partial}=I-H,\qquad GH=HG=0.
$$
For any form $u$, expand the Laplacian term to obtain
$$
u=Hu+\bar\partial\bar\partial^*Gu+\bar\partial^*\bar\partial Gu.
$$
Harmonic forms are both closed and coclosed, since $(\Delta_{\bar\partial}v,v)=\|\bar\partial v\|^2+\|\bar\partial^*v\|^2$. They are orthogonal to both other terms; the two images are orthogonal because $(\bar\partial a,\bar\partial^*b)=(\bar\partial^2a,b)=0$. This proves the <Dolbeault Hodge decomposition>
$$
\boxed{A^{p,q}=\mathcal H^{p,q}_{\bar\partial}\oplus\bar\partial A^{p,q-1}\oplus\bar\partial^*A^{p,q+1}.}
$$
The closed forms are exactly the first two summands: if $\bar\partial u=0$, then $u$ is orthogonal to the adjoint image, and conversely the first two summands are closed. It follows that every <Dolbeault cohomology> class has a unique harmonic representative. Negative or out-of-range bidegrees mean zero spaces. Compactness and the absence of boundary are necessary for this stated global decomposition.
Now assume a <Kähler metric>. Set $L=\omega\wedge-$ and $\Lambda=L^*$. The permitted <Kähler identities>, stated with signs, are
$$
[\Lambda,\partial]=i\bar\partial^*,\qquad[\Lambda,\bar\partial]=-i\partial^*,
$$
with ordinary commutators because $\Lambda$ has even degree. Thus $\bar\partial^*=-i[\Lambda,\partial]$, and
$$
\partial\bar\partial^*+\bar\partial^*\partial=-i\{\partial(\Lambda\partial-\partial\Lambda)+(\Lambda\partial-\partial\Lambda)\partial\}=0
$$
by $\partial^2=0$. The companion identity gives $\bar\partial\partial^*+\partial^*\bar\partial=0$ in the same way.
Define $\Delta_d=dd^*+d^*d$ and $\Delta_\partial=\partial\partial^*+\partial^*\partial$. Expanding $d=\partial+\bar\partial$ and $d^*=\partial^*+\bar\partial^*$ makes the two mixed anticommutators vanish, so $\Delta_d=\Delta_\partial+\Delta_{\bar\partial}$. To prove equality of the latter two, substitute the same commutator formulas and use $\partial\bar\partial=-\bar\partial\partial$:
$$
\begin{aligned}
\Delta_\partial&=i(\partial\Lambda\bar\partial-\partial\bar\partial\Lambda-\Lambda\partial\bar\partial-\bar\partial\Lambda\partial),\\
\Delta_{\bar\partial}&=-i(\bar\partial\Lambda\partial+\partial\bar\partial\Lambda+\Lambda\partial\bar\partial-\partial\Lambda\bar\partial).
\end{aligned}
$$
The displayed operators are identical. Therefore the <Kähler Laplacian identity> is
$$
\boxed{\Delta_d=\Delta_\partial+\Delta_{\bar\partial}=2\Delta_\partial=2\Delta_{\bar\partial}.}
$$
Let $\eta$ be a $\bar\partial$-exact form. It is closed and orthogonal to harmonic forms. The <Dolbeault Hodge decomposition>, or the Green identity with $\alpha=G\eta$, therefore gives
$$
\boxed{\eta=\bar\partial\bar\partial^*\alpha,\qquad\alpha\in A^{p,q}.}
$$
For the Green construction, $\bar\partial\alpha=0$ follows since $G$ commutes with $\bar\partial$; equivalently the coexact component of a closed form vanishes. This yields exactly the stated factorization.
If also $\partial\eta=0$, put $\beta=\bar\partial^*\alpha$ and $\xi=\partial\beta$. The identities already proved give
$$
\bar\partial\xi=-\partial\bar\partial\beta=-\partial\eta=0,\qquad \bar\partial^*\xi=-\partial(\bar\partial^*)^2\alpha=0.
$$
Thus $\xi$ is harmonic for $\Delta_{\bar\partial}$ and hence for $\Delta_\partial$. But it is $\partial$-exact, so it is orthogonal to the harmonic summand in the <Hodge decomposition theorem>. Consequently $\xi=0$, proving that \b[$\beta=\bar\partial^*\alpha$ is $\partial$-closed].
Furthermore $\beta$ is orthogonal to the common harmonic space, since $(\beta,h)=(\alpha,\bar\partial h)=0$. Its $\partial$-Hodge decomposition then has neither a harmonic term nor a coexact term: closedness kills the coexact term. Therefore $\beta=\partial\chi$ for a form $\chi\in A^{p-1,q-1}$. It follows that
$$
\eta=\bar\partial\partial\chi=-\partial\bar\partial\chi,\qquad \boxed{\eta=\partial\bar\partial\phi,\quad\phi=-\chi\in A^{p-1,q-1}.}
$$
This proves the requested <ddbar lemma>, including the intermediate harmonicity and orthogonality steps. The hypotheses $p,q\geq1$ ensure that the potential has a valid bidegree.
Finally $\eta=\omega_2-\omega_1$ is a real closed $(1,1)$ form and is $d$-exact. Integration by parts makes it orthogonal to every $d$-harmonic form. By the <Kähler Laplacian identity> it is also orthogonal to the Dolbeault harmonic space; because it is $\bar\partial$-closed, the <Dolbeault Hodge decomposition> makes it $\bar\partial$-exact. Apply the proved <ddbar lemma> to get $\eta=\partial\bar\partial\phi$ for a smooth complex-valued function. Reality gives $\eta=-\partial\bar\partial\overline\phi$, so averaging yields
$$
\eta=\frac12\partial\bar\partial(\phi-\overline\phi)=i\partial\bar\partial\operatorname{Im}\phi.
$$
With $f=\operatorname{Im}\phi$,
$$
\boxed{\omega_2=\omega_1+i\partial\bar\partial f,\qquad f\in C^\infty(M,\mathbb R).}
$$
The potential is defined up to a real constant on each connected component. This is a statement about cohomologous <Kähler forms>, not about an arbitrary pair of such forms.
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