I'm not sure this answers your question since I think that "by hands" might have different meanings, but it seems to me that what you are asking for is precisely the content of Proposition 2.2 on page 14 of the paper of Roberts-Willerton you mentioned. 

Let me explain what they do. The Atiyah class of a vector bundle $E$ can be represented by a Dolbeault cocycle: pick a smooth hermitian connection in $E$ and consider its curvature form $R$, which one views as a Dolbeault $1$-cocycle with values in $\Omega^1_X\otimes End(E)$. When $E=T_X$, Proposition 2.2 on page 14 of Roberts-Willerton's paper says that the skew-symmetrization of $R$ is a coboundary. More precisely, $R-R^{op}=\overline\partial(T^{1,0})$, where $T^{1,0}$ is the $(1,0)$ component of the torsion of the above hermitian connection. 

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In any case, here is another proof that uses exact sequences... which you could consider being both conceptual and "by hands". 

The Atiyah class of $E$ is the class of the following exact sequence: 
$$
0\to E\to Diff^{\leq 1}_X(E)\to T_X\otimes E\to 0 \qquad(1)
$$
where $Diff^{\leq 1}_X(E)$ is the sheaf of $E$-valued differential operators of degree $\leq1$. 

In the case $E=T_X$ this gives 
$$
0\to T_X\to Diff^{\leq 1}_X(T_X)\to T_X\otimes T_X\to 0\qquad(2)
$$
and we have a morphism from it to the following one: 
$$
0\to T_X\to Diff^{+,\leq 2}_X\to S^2(T_X)\to 0\qquad(3)
$$
where $Diff^{+,\leq 2}_X$ is the sheaf of differential operators (with values in $\mathcal O_X$) of degree $\leq2$ and vanishing on constants. Moreover one can prove that  
$$
T_X=\ker\left(T_X\oplus Diff^{\leq 1}_X(T_X)\to Diff^{+,\leq 2}_X\right)\qquad(4)
$$
Hence the result. 

From a more geometric point of view we have that: 

+ Exact sequence (1), resp. (2), splits if and only if there exists a holomorphic connection in $E$, resp. $T_X$. 

+ Exact sequence (3) splits if and only if there exists a torsion free holomorphic connection in $T_X$. 

+ (4) insures that (2) splits if and only if (3) does.