Since an elliptic curve is determined by its double branched cover to $P^1$, the coarse space $\overline{M}_{1,1}$ is isomorphic to $\overline{M}_{0,4}/S_3$ where the quotient by $S_3$ is because only one of the branched points is marked. Thus $\overline{M}_{1,2} $ is isomorphic to $ \overline{M}_{0,5}/S_3$, at least on the level of coarse spaces.

$\overline{M}_{0,5}$ is famously the blowup of $CP^2$ at four points and the map from $\overline{M}_{0,5}$ to $\overline{M}_{0,4}$ is the family given by the linear system of conics through the 4 points. If I take those 4 points in $CP^2$ to be (1:0:0), (0:1:0), (0:0:1), (1:1:1), then I can perform the quotient pretty explicitly --- it is induced by the permutation of the homogeneous coordinates $(x,y,z)$. This quotient is proj of the ring of symmetric functions which is generated by $u=x+y+z$, $v=xy+xz+yz$, and $w=xyz$. This is the weighted projective space $CP(1,2,3)$ and the orbits of the 4 points in $CP^2$ become two points $(1:0:0)$ and $(3:2:1)$.

So I think (if all my above logic is correct) that $\overline{M}_{1,2}$ is isomorphic (as coarse spaces) to the blowup of the weighted projective space $CP(1,2,3)$ at the two points $(1:0:0)$ and $(3:2:1)$. Note that $CP(1,2,3)$ has only two singular points at (0:1:0) and (0:0:1) (of type $A_1$ and $A_2$) which are away from the blowup points.

The same kind of argument should work on $\overline{M}_{2,1}$ since by similar logic the coarse space of is isomorphic to $\overline{M}_{0,7}/S_6$ and $\overline{M}_{0,7}$ has a description as a blowup of $CP^4$ (with the ruling given by a pencil of rational normal curves).