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 $t = 0$.
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 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 $(0:1)$.