$$\partial_{\mu} \frac{\partial(\partial_{\alpha}A_{\alpha})^2}{\partial(\partial_{\mu}A_{\nu})} = \partial_{\mu}\left[2(\partial_{\alpha}A_{\alpha})\frac{\partial(\partial_{\beta}A_{\gamma})}{\partial(\partial_{\mu}A_{\nu})} g_{\beta\gamma}\right]$$
I assume the metric comes from differentiating a vector wrt to another and chain rule is applied, but I can't see how exactly these two sides are equal.
As mentioned in the OP's link Schwartz (in his QFT book) doesn't keep track of the index placement on tensor objects that might obscure the structure a little.
However, ignoring the first derivative $\partial_\mu$ (because it is not relevant for the answer) the starting point is to distinguish between covariant and contravariant indices such that $$ (\partial_\alpha A_\alpha)^2 \to (\partial_\alpha A^\alpha)^2=(\partial_\alpha A^\alpha)(\partial_\beta A^\beta)\ .$$
Applying the product rule we see $$\frac{\partial (\partial_\alpha A^\alpha)(\partial_\beta A^\beta)}{\partial (\partial_\mu A_\nu)}= \frac{\partial (\partial_\alpha A^\alpha)}{\partial (\partial_\mu A_\nu)}(\partial_\beta A^\beta)+(\partial_\alpha A^\alpha)\frac{\partial(\partial_\beta A^\beta)}{\partial (\partial_\mu A_\nu)}\ .$$
Since the indices must be at the right position in order to perform a derivative we can use the metric to write $\partial_\alpha A^\alpha= g^{\alpha\gamma}\partial_\alpha A_\gamma$ similar for the second expression.
Hence, the Euler-Lagrange variation involves a metric $$\frac{\partial (\partial_\alpha A^\alpha)}{\partial (\partial_\mu A_\nu)}=\frac{\partial (\partial_\alpha A_\gamma)}{\partial (\partial_\mu A_\nu)}g^{\alpha\gamma}\ .$$
Collecting all pieces and replacing the labeling of dummy indices $\alpha \leftrightarrow \beta$ this leads eventually to the result in the brackets $$\frac{\partial (\partial_\alpha A^\alpha)(\partial_\beta A^\beta)}{\partial (\partial_\mu A_\nu)} = (\partial_\beta A^\beta)\frac{\partial (\partial_\alpha A_\gamma)}{\partial (\partial_\mu A_\nu)}g^{\alpha\gamma}+(\partial_\alpha A^\alpha)\frac{\partial(\partial_\beta A_\gamma)}{\partial (\partial_\mu A_\nu)}g^{\beta \gamma}\\ =(\partial_\alpha A^\alpha)\frac{\partial (\partial_\beta A_\gamma)}{\partial (\partial_\mu A_\nu)}g^{\beta\gamma}+(\partial_\alpha A^\alpha)\frac{\partial(\partial_\beta A_\gamma)}{\partial (\partial_\mu A_\nu)}g^{\beta \gamma}\\ =2 (\partial_\alpha A^\alpha)\frac{\partial (\partial_\beta A_\gamma)}{\partial (\partial_\mu A_\nu)}g^{\beta\gamma}\ .$$
