[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 $H^2(X,T_X)$.[/step]

[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: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]