In this answer it is said that $\bar{M}_{2,1}\cong \bar{M}_{0,7}/S_6$. However, I cannot see this. Given a curve of genus $2$ and a marked point, quotienting by the involution surely gives a rational curve with a divisor of degree $6$ together with a distinct marked point. However, going the other way, from a rational curve with a divisor of degree $6$ with a distinct rational point we get a genus $2$ curve, but we (generally) have two choices to lift the distinguished marking. In other words, that map looks more like a (ramified) double cover that an isomorphism.

Where is my reasoning wrong? Is there a way to fix this (either my wrong reasoning, or Jim's wrong reasoning)? The objective is still to find a presentation of $\bar{M}_{2,1}$, as similar as possible to the one proposed.

Note: I agree with his proof in the case $\bar{M}_{1,2}$.