I'm reading a paper, where they consider the hyperelliptic curve $$C : y^2 = x^2(x^2-1)^2 - t^2$$ over the ring $k[[t]]$, where $k$ is a field. Let $K = k((t))$, then they say:

"The stable reduction of $C_K$ is the union of two copies of $\mathbb{P}^1$ intersecting at 3 points"

This seems reasonable to me, since at $t = 0$, you get $y^2 = x^2(x^2-1)^2$, so an affine patch is given by $$\text{Spec }k[x,y]/(y-x(x^2-1)(y+x(x^2-1))$$ which is visibly two genus 0 curves intersecting at the points $(1,0),(0,0),(-1,0)$. However, the usual projectivization of the equation for $C$ yields a curve with a unique point at $\infty$ which is singular, but I assume that when they speak of $C_K$, they must be referring to the smooth model of $C_K$.

Though, I'm not sure how to carry out the computation to see that the two components of the smooth model of $C_K$ do not intersect "at infinity".

Next, they say:

"Another stable degeneration of $C_K$ is the union of two curves of genus 1 intersecting transversally at a point"

This part I don't understand. Of course if we adjoin $\sqrt{y}$, then the same computation as above will yield two nodal cubics (in $x,\sqrt{y}$) of the form $(\sqrt{y})^2 = \pm x(x^2-1)$, but I don't see how we can view this as a degeneration of $C_K$. What am I missing?