[proofplan]
We first check that contraction with the fixed holomorphic volume form sends a Kodaira-Spencer representative to a well-defined Dolbeault class in $H^{n-1,1}(X)$. Then we invoke the general infinitesimal period-map formula, which identifies the first-order variation of the Hodge filtration with contraction by the Kodaira-Spencer class. Specializing that formula to the summand $H^{n,0}(X)$ gives exactly the displayed map from $H^1(X,T_X)$ to $\operatorname{Hom}(H^{n,0}(X),H^{n-1,1}(X))$.
[/proofplan]
[step:Show that contraction with $\Omega$ gives a well-defined class in $H^{n-1,1}(X)$]
Let $\varphi\in A^{0,1}(X,T_X)$ be a smooth $T_X$-valued $(0,1)$-form satisfying $\bar\partial\varphi=0$. Define the contraction map
\begin{align*}
C_\Omega:A^{0,1}(X,T_X)&\to A^{n-1,1}(X)
\end{align*}
by
\begin{align*}
C_\Omega(\psi):=\psi\lrcorner\Omega
\end{align*}
for every $\psi\in A^{0,1}(X,T_X)$. Thus $C_\Omega(\varphi)=\varphi\lrcorner\Omega\in A^{n-1,1}(X)$.
Since $\Omega$ is holomorphic, $\bar\partial\Omega=0$. The Dolbeault operator on contractions satisfies the graded Leibniz rule
\begin{align*}
\bar\partial(\varphi\lrcorner\Omega)=(\bar\partial\varphi)\lrcorner\Omega-\varphi\lrcorner(\bar\partial\Omega).
\end{align*}
Both terms on the right vanish, so $\bar\partial(\varphi\lrcorner\Omega)=0$. Thus $\varphi\lrcorner\Omega$ defines a Dolbeault cohomology class
\begin{align*}
[\varphi\lrcorner\Omega]\in H^1(X,\Omega_X^{n-1})\cong H^{n-1,1}(X).
\end{align*}
If $\varphi'$ is another representative of the same class in $H^1(X,T_X)$, then there exists a smooth section $\xi\in A^{0,0}(X,T_X)$ such that
\begin{align*}
\varphi'=\varphi+\bar\partial\xi.
\end{align*}
Applying the same graded Leibniz rule to $\xi\lrcorner\Omega$ gives
\begin{align*}
(\bar\partial\xi)\lrcorner\Omega=\bar\partial(\xi\lrcorner\Omega),
\end{align*}
because $\bar\partial\Omega=0$. Therefore
\begin{align*}
\varphi'\lrcorner\Omega-\varphi\lrcorner\Omega=\bar\partial(\xi\lrcorner\Omega),
\end{align*}
so the Dolbeault class $[\varphi\lrcorner\Omega]$ depends only on $[\varphi]\in H^1(X,T_X)$.
[guided]
The first point is that the displayed formula has to make sense in cohomology, not just on representatives. Let $\varphi\in A^{0,1}(X,T_X)$ be a smooth $T_X$-valued $(0,1)$-form with $\bar\partial\varphi=0$. Contracting the tangent-vector component into the holomorphic $n$-form $\Omega$ defines a map
\begin{align*}
C_\Omega:A^{0,1}(X,T_X)&\to A^{n-1,1}(X)
\end{align*}
by
\begin{align*}
C_\Omega(\psi):=\psi\lrcorner\Omega
\end{align*}
for every $\psi\in A^{0,1}(X,T_X)$. In particular, $C_\Omega(\varphi)=\varphi\lrcorner\Omega$ is an ordinary $(n-1,1)$-form.
We must check that this form is $\bar\partial$-closed. The compatibility of $\bar\partial$ with contraction gives the graded identity
\begin{align*}
\bar\partial(\varphi\lrcorner\Omega)=(\bar\partial\varphi)\lrcorner\Omega-\varphi\lrcorner(\bar\partial\Omega).
\end{align*}
The first term vanishes because $\varphi$ represents a Dolbeault cohomology class, hence $\bar\partial\varphi=0$. The second term vanishes because $\Omega$ is holomorphic, hence $\bar\partial\Omega=0$. Therefore
\begin{align*}
\bar\partial(\varphi\lrcorner\Omega)=0.
\end{align*}
Thus $\varphi\lrcorner\Omega$ determines a class in $H^1(X,\Omega_X^{n-1})$, which is the same Hodge summand as $H^{n-1,1}(X)$ for the compact Kähler manifold $X$.
Now suppose we change the representative of the Kodaira-Spencer class. If $\varphi'$ and $\varphi$ represent the same class in $H^1(X,T_X)$, then their difference is $\bar\partial$-exact, so there is a smooth section $\xi\in A^{0,0}(X,T_X)$ satisfying
\begin{align*}
\varphi'=\varphi+\bar\partial\xi.
\end{align*}
Contracting with $\Omega$ gives
\begin{align*}
\varphi'\lrcorner\Omega-\varphi\lrcorner\Omega=(\bar\partial\xi)\lrcorner\Omega.
\end{align*}
Because $\bar\partial\Omega=0$, the graded Leibniz rule applied to $\xi\lrcorner\Omega$ gives
\begin{align*}
(\bar\partial\xi)\lrcorner\Omega=\bar\partial(\xi\lrcorner\Omega).
\end{align*}
So the difference between the two contracted forms is $\bar\partial$-exact. Hence the cohomology class $[\varphi\lrcorner\Omega]\in H^{n-1,1}(X)$ is independent of the chosen representative $\varphi$.
[/guided]
[/step]
[step:Apply the general infinitesimal period-map formula to the top Hodge summand]
Let $v\in T_{[X]}\mathcal M$ be a tangent vector, and let
\begin{align*}
\operatorname{KS}_{[X]}:T_{[X]}\mathcal M\to H^1(X,T_X)
\end{align*}
denote the Kodaira-Spencer map of the family $\pi:\mathcal X\to\mathcal M$ at the fibre $X=\pi^{-1}([X])$. By the Kodaira-Spencer identification, write
\begin{align*}
\operatorname{KS}_{[X]}(v)=[\varphi]
\end{align*}
for a $\bar\partial$-closed representative $\varphi\in A^{0,1}(X,T_X)$.
By the [[Kodaira-Spencer Stability Theorem for Kähler Manifolds](/theorems/9126)][citetheorem:9126], after shrinking $\mathcal M$ around $[X]$, every fibre of the smooth proper holomorphic submersion $\pi:\mathcal X\to\mathcal M$ is a compact Kähler manifold. The [[Formula for the Infinitesimal Period Map](/theorems/9131)][citetheorem:9131] therefore applies to this family with the chosen local marking of the flat local system $R^n\pi_*\mathbb C$. We use the sign convention for the Kodaira-Spencer class and contraction under which that formula reads $[\varphi]\mapsto(\alpha\mapsto[\varphi\lrcorner\alpha])$. Specializing the theorem to weight $n$ and to the Hodge summand $H^{n,0}(X)$ gives
\begin{align*}
d\mathcal P_{[X]}(v)(\alpha)=[\varphi\lrcorner\alpha]
\end{align*}
for every element $\alpha\in H^{n,0}(X)$, viewed as the Dolbeault class of a holomorphic $n$-form $\alpha\in H^0(X,K_X)$.
Taking $\alpha=\Omega$ yields
\begin{align*}
d\mathcal P_{[X]}(v)(\Omega)=[\varphi\lrcorner\Omega]\in H^{n-1,1}(X).
\end{align*}
Since $v$ corresponds to $[\varphi]$ under the Kodaira-Spencer identification, this is precisely
\begin{align*}
d\mathcal P_{[X]}([\varphi])(\Omega)=[\varphi\lrcorner\Omega].
\end{align*}
[/step]
[step:Identify the target with $\operatorname{Hom}(H^{n,0}(X),H^{n-1,1}(X))$]
By [Griffiths transversality](/theorems/9129), recorded in [Griffiths Transversality][citetheorem:9129], the derivative of the period map sends the top Hodge filtration piece $F^nH^n(X,\mathbb C)=H^{n,0}(X)$ into the quotient
\begin{align*}
F^{n-1}H^n(X,\mathbb C)/F^nH^n(X,\mathbb C).
\end{align*}
For a compact Kähler manifold, the [Hodge decomposition](/theorems/2745) identifies this quotient with $H^{n-1,1}(X)$. Therefore the derivative restricts to a [linear map](/page/Linear%20Map)
\begin{align*}
d\mathcal P_{[X]}:H^1(X,T_X)\to \operatorname{Hom}\bigl(H^{n,0}(X),H^{n-1,1}(X)\bigr).
\end{align*}
The form $\Omega$ generates the one-dimensional space $H^{n,0}(X)$: if $\alpha\in H^0(X,K_X)$, then the nowhere-vanishing property of $\Omega$ gives a [holomorphic function](/page/Holomorphic%20Function) $f:X\to\mathbb C$ with $\alpha=f\Omega$, and compact connectedness of $X$ forces $f$ to be constant. The preceding step computes the derivative on this generator, and the first step shows that the resulting class depends only on $[\varphi]$. Hence the derivative of the period map is exactly the claimed map.
[/step]
