Let $F_i:\mathbb R\to[0,1]$ be a distribution function$^1$ and $$F_i^{-1}(t):=\inf\left\{x\in\mathbb R:F_i(x)\ge t\right\}\;\;\;\text{for }t\in[0,1].$$

I've got a computer program where only $F_i^{-1}$ (not $F_i$) is directly available. Assume $t_1\in[0,1]$ and $x:=F_1^{-1}(t_1)\in F_2^{-1}([0,1])$. I need to compute a $t_2\in[0,1]$ such that $x=F_2^{-1}(t_2)$. How can I do this?

First of all, we know that $$\left\{t_2\in[0,1]:F_2^{-1}(t_2)=x\right\}=\begin{cases}[F_2(x-),F_2(x)]&\text{, if }F_2\text{ is continuous at }x\text{ or }\forall y<x:F_2(y)<F_2(x-)\\(F_2(x-),F_2(x)]&\text{, otherwise}.\end{cases}$$

Now, I've read (here in section 6.1) the following, but can't really make sense of it: Assuming that $F_2^{-1}([a,b])=\{x\}$ for some $0\le a\le b\le 1$ we can find $t_2$ by sampling $u$ with uniform distribution on $[0,1]$ and set $t_2:=a+(b-a)u$. Why does this work? And if it works, how can we apply it in practice (i.e. how do we find $a,b$)?

Remark: I think this book page (above Example 3.31) is related.

$^1$ i.e. $F_i$ is right-continuous and nondecreasing with $F(-\infty):=\lim_{x\to-\infty}F(x)=0$ and $F(\infty):=\lim_{x\to\infty}F(x)=1$.