[proofplan]
We expand the Maurer-Cartan equation for a second-order formal deformation and compare coefficients of the formal parameter $t$. The coefficient of $t$ is exactly the given condition $\bar\partial\varphi_1=0$. The coefficient of $t^2$ forces
\begin{align*}
\frac{1}{2}[\varphi_1,\varphi_1]
\end{align*}
to be the negative of a $\bar\partial$-boundary, so its Dolbeault cohomology class must vanish. We also record why this expression is a legitimate degree-two cohomology class: the Kodaira-Spencer bracket is compatible with $\bar\partial$, and $\bar\partial\varphi_1=0$.
[/proofplan]
[step:Identify the degree of the primary obstruction term]
Let
\begin{align*}
[\cdot,\cdot]:A^{0,p}(X,T_X)\times A^{0,q}(X,T_X)\to A^{0,p+q}(X,T_X)
\end{align*}
denote the Kodaira-Spencer bracket on $T_X$-valued forms. Since $\varphi_1\in A^{0,1}(X,T_X)$, the bracket $[\varphi_1,\varphi_1]$ lies in $A^{0,2}(X,T_X)$. Hence $\frac{1}{2}[\varphi_1,\varphi_1]$ has the correct bidegree to define a class in $H^2(X,T_X)$ once it is $\bar\partial$-closed.
[/step]
[step:Check that the obstruction term is $\bar\partial$-closed]
The defining compatibility identity for the Kodaira-Spencer bracket says that the Dolbeault operator is a graded derivation: for $\alpha\in A^{0,p}(X,T_X)$ and $\beta\in A^{0,q}(X,T_X)$,
\begin{align*}
\bar\partial[\alpha,\beta]=[\bar\partial\alpha,\beta]+(-1)^p[\alpha,\bar\partial\beta].
\end{align*}
Applying this with $\alpha=\beta=\varphi_1$ and $p=1$, we obtain
\begin{align*}
\bar\partial[\varphi_1,\varphi_1]=[\bar\partial\varphi_1,\varphi_1]-[\varphi_1,\bar\partial\varphi_1].
\end{align*}
Because $\bar\partial\varphi_1=0$, both terms on the right-hand side vanish. Therefore
\begin{align*}
\bar\partial\left(\frac{1}{2}[\varphi_1,\varphi_1]\right)=0.
\end{align*}
Thus $\frac{1}{2}[\varphi_1,\varphi_1]$ determines a Dolbeault cohomology class in $H^2(X,T_X)$.
[guided]
The expression $\left[\frac{1}{2}[\varphi_1,\varphi_1]\right]$ only makes sense as a Dolbeault cohomology class if the representative is $\bar\partial$-closed. We verify that point directly.
The defining compatibility identity for the Kodaira-Spencer bracket says that $\bar\partial$ is a graded derivation:
\begin{align*}
\bar\partial[\alpha,\beta]=[\bar\partial\alpha,\beta]+(-1)^p[\alpha,\bar\partial\beta]
\end{align*}
for every $\alpha\in A^{0,p}(X,T_X)$ and every $\beta\in A^{0,q}(X,T_X)$. In our case $\alpha=\beta=\varphi_1$ and $p=1$, so
\begin{align*}
\bar\partial[\varphi_1,\varphi_1]=[\bar\partial\varphi_1,\varphi_1]-[\varphi_1,\bar\partial\varphi_1].
\end{align*}
The hypothesis says that $\varphi_1$ is $\bar\partial$-closed, namely $\bar\partial\varphi_1=0$. Substituting this into the displayed identity gives
\begin{align*}
\bar\partial[\varphi_1,\varphi_1]=0.
\end{align*}
Multiplying by the scalar $\frac{1}{2}$ preserves $\bar\partial$-closedness, so
\begin{align*}
\bar\partial\left(\frac{1}{2}[\varphi_1,\varphi_1]\right)=0.
\end{align*}
Therefore $\frac{1}{2}[\varphi_1,\varphi_1]\in A^{0,2}(X,T_X)$ is a valid representative of a class in $H^2(X,T_X)$.
[/guided]
[/step]
[step:Expand the Maurer-Cartan equation through order $t^2$]
By the meaning of extension to a second-order deformation, there exists $\varphi_2\in A^{0,1}(X,T_X)$ such that
\begin{align*}
\varphi(t)=t\varphi_1+t^2\varphi_2
\end{align*}
satisfies
\begin{align*}
\bar\partial\varphi(t)+\frac{1}{2}[\varphi(t),\varphi(t)]=0
\end{align*}
modulo $t^3$. Since $\bar\partial$ is linear,
\begin{align*}
\bar\partial\varphi(t)=t\bar\partial\varphi_1+t^2\bar\partial\varphi_2.
\end{align*}
Since the Kodaira-Spencer bracket is bilinear, the terms of degree less than $3$ in $[\varphi(t),\varphi(t)]$ consist only of $t^2[\varphi_1,\varphi_1]$. Therefore, modulo $t^3$, the Maurer-Cartan equation becomes
\begin{align*}
t\bar\partial\varphi_1+t^2\left(\bar\partial\varphi_2+\frac{1}{2}[\varphi_1,\varphi_1]\right)=0
\end{align*}
modulo $t^3$.
[/step]
[step:Compare the $t^2$ coefficient and conclude exactness]
Equality modulo $t^3$ implies that the coefficient of $t^2$ vanishes. Hence
\begin{align*}
\bar\partial\varphi_2+\frac{1}{2}[\varphi_1,\varphi_1]=0.
\end{align*}
Equivalently,
\begin{align*}
\frac{1}{2}[\varphi_1,\varphi_1]=-\bar\partial\varphi_2.
\end{align*}
Thus $\frac{1}{2}[\varphi_1,\varphi_1]$ is $\bar\partial$-exact in $A^{0,2}(X,T_X)$. Its Dolbeault cohomology class is therefore zero:
\begin{align*}
\left[\frac{1}{2}[\varphi_1,\varphi_1]\right]=0 \quad \text{in } H^2(X,T_X).
\end{align*}
This proves the necessary condition for the first-order class $[\varphi_1]$ to extend to second order.
[/step]
