I don't know any literature for how to do this, but I think I can do this example by hand in an ad hoc way.
As you say you have to blow up $\mathbb C \times \mathbb P^1$ in the two points you gave, so the special fiber becomes a chain of three $\mathbb P^1$s, with a marked point on each component. On the middle one of these components you get the constant map to $(0:1)$ when $t = 0$. On the other two you will have degrees 1 and 2.
The evaluation maps will send $(1:0) \mapsto (0:1)$ for all $t$, [which is a good sanity check since this will be the marked point on the middle component which maps constantly to $(0:1)$], and will map $(1:1) \mapsto (1:0)$ and $(1:-1) \mapsto (1:0)$ for $t \neq 0$. But then this must be true in the limit too, so on both of the two new components we will have two special points, one which meets the middle component and must go to $(0:1)$ and one marking which must go to $(1:0)$.
Now we consider the branch points of the stable map. There are four branch points counted with multiplicity by Riemann-Hurwitz, and one computes that $(0:1)$ is a double branch point, $(1:0)$ is simple, and the third one has the form $(t\alpha:1)$ for some nonzero $\alpha$. So in the limit we will have two branch points, one of multiplicity three and one of multiplicity one.
For a stable curve, the image of a node should be always counted as a branch point with multiplicity one, which means that the point $(0:1)$ needs one more ramification point over it.
Now by looking at the equation one is led to think that the map should have degree 1 on the component given by blowing up at $(1:1)$ and degree 2 on the component given by blowing up at $(1:-1)$. I think one way of seeing this is that the point $(1:-1)$ is a ramification point of the stable map for $t \neq 0$, hence this should be true in the limit, which would be impossible if it were on a degree 1 component. So the map on the $(1:1)$-component is an isomorphism, and the map on the $(1:-1)$-component is the unique branched double cover of $\mathbb P^1$ branched over $(1:0)$ and $(0:1)$.