One thing that confused me about Francesco's answer was how to actually construct the branch covers $f_k:Y_k\to S$ which are branched over the vertex and a given curve. Since I was sheepish enough not to ask, perhaps somebody else (maybe future me) will benefit from a description.

Let $g(x,y,z)$ be a polynomial which does not vanish at the origin. We then have two interesting degree 2 maps to $S=Spec(k[a^2,ab,b^2])$:

• $\mathbb A^2\to S$, corresponding to the inclusion $k[a^2,ab,b^2]\to k[a,b]$. Think of $S$ as $\mathbb A^2/\mu_2$, where $\mu_2$ acts by $(a,b)\mapsto (-a,-b)$. This is branched only over the vertex, since $(0,0)$ is the only point with a non-trivial stabilizer.
• $S[\sqrt{g}]\to S$ (almost certainly non-standard notation since I just made it up), corresponding to the inclusion of rings $k[a^2,ab,b^2]\to k[a^2,ab,b^2,\sqrt{g(a^2,ab,b^2)}]$. Think of $S$ as $S[\sqrt g]/\mu_2$, where $\mu_2$ acts by $\sqrt g\mapsto -\sqrt g$. This is branched over the vanishing locus of $g$, since that's exactly where you have non-trivial stabilizer.

We can then define a sort of common refinement, $\tilde Y=Spec(k[a,b,\sqrt{g(a^2,ab,b^2)}]$, which has an action of $\mu_2\times \mu_2$. Quotienting by the first $\mu_2$ gives us $S[\sqrt g]$. Quotienting by the second $\mu_2$ gives us $\mathbb A^2$. Quotienting by both gives you $S$. Define $Y$ as the quotient by the diagonal $\mu_2$ action, $(a,b,\sqrt g)\mapsto (-a,-b,-\sqrt g)$.^† This action is free since $g(0,0,0)\neq 0$, so $\tilde Y\to Y$ is actually an etale cover. If $V(g)\cap S$ is smooth, $\tilde Y$ is smooth, so $Y$ is smooth. We have a remaining $\mu_2$ action on $Y$ with $Y/\mu_2 = S$.

$$\begin{array}{cccccc} & & \tilde Y\\ & \swarrow & \downarrow & \searrow\\ \mathbb A^2 & & Y & & S[\sqrt g]\\ & \searrow & \downarrow & \swarrow \\ & & S \end{array}$$

^† You can very explicitly describe the ring of invariants under this action. $Y$ is the spectrum of $k[a^2,ab,b^2,a\sqrt g,b\sqrt g]$. The $\mu_2$ action on $Y$ is $(a^2,ab,b^2,a\sqrt g,b\sqrt g)\mapsto (a^2,ab,b^2,-a\sqrt g,-b\sqrt g)$.

One thing that confused me about Francesco's answer was how to actually construct the branch covers $f_k:Y_k\to S$ which are branched over the vertex and a given curve. Since I was sheepish enough not to ask, perhaps somebody else (maybe future me) will benefit from a description.

Let $g(x,y,z)$ be a polynomial which does not vanish at the origin. We then have two interesting degree 2 maps to $S=Spec(k[a^2,ab,b^2])$:

• $\mathbb A^2\to S$, corresponding to the inclusion $k[a^2,ab,b^2]\to k[a,b]$. Think of $S$ as $\mathbb A^2/\mu_2$, where $\mu_2$ acts by $(a,b)\mapsto (-a,-b)$. This is branched only over the vertex, since $(0,0)$ is the only point with a non-trivial stabilizer.
• $S[\sqrt{g}]\to S$ (almost certainly non-standard notation since I just made it up), corresponding to the inclusion of rings $k[a^2,ab,b^2]\to k[a^2,ab,b^2,\sqrt{g(a^2,ab,b^2)}]$. Think of $S$ as $S[\sqrt g]/\mu_2$, where $\mu_2$ acts by $\sqrt g\mapsto -\sqrt g$. This is branched over the vanishing locus of $g$, since that's exactly where you have non-trivial stabilizer.

We can then define a sort of common refinement, $\tilde Y=Spec(k[a,b,\sqrt{g(a^2,ab,b^2)}]$, which has an action of $\mu_2\times \mu_2$. Quotienting by the first $\mu_2$ gives us $S[\sqrt g]$. Quotienting by the second $\mu_2$ gives us $\mathbb A^2$. Quotienting by both gives you $S$. Define $Y$ as the quotient by the diagonal $\mu_2$ action, $(a,b,\sqrt g)\mapsto (-a,-b,-\sqrt g)$.^† This action is free since $g(0,0,0)\neq 0$, so $\tilde Y\to Y$ is actually an etale cover. If $V(g)\cap S$ is smooth, $\tilde Y$ is smooth, so $Y$ is smooth. We have a remaining $\mu_2$ action on $Y$ with $Y/\mu_2 = S$.

$$\begin{array}{cccccc} & & \tilde Y\\ & \swarrow & \downarrow & \searrow\\ \mathbb A^2 & & Y & & S[\sqrt g]\\ & \searrow & \downarrow & \swarrow \\ & & S \end{array}$$

^† You can very explicitly describe the ring of invariants under this action. $Y$ is the spectrum of $k[a^2,ab,b^2,a\sqrt g,b\sqrt g]$. The $\mu_2$ action on $Y$ is $(a^2,ab,b^2,a\sqrt g,b\sqrt g)\mapsto (a^2,ab,b^2,-a\sqrt g,-b\sqrt g)$.