I need to ask, where does the number $ b/M $ come from in the conditions required for there to be existence of solution
When they say:: $\alpha = \min( a, b/M )$
Theorem: $R=\{(x,y):|x−x_0|≤a,|y−y_0|≤b\}$. Then the IVP $y'=f(x,y)$, $y(x_0)=y_0$ has an unique solution in the interval $|x−x_0|≤α=\min(a,b/M)$ where $M=\max_{(x,y)∈ R}|f(x,y)|$.
Note that apriori the function $(t,x)\mapsto f(t,x)$ is defined around $(t_0,x_0)$, at least on a closed rectangle $[t_0-a,t_0+a]\times [x_0-b, x_0+b]$. You consider the Picard iterations $x_0(t)\equiv x_0$, $x_{n+1}(t)=x_0+\int_{t_0}^t f(t,x_n(s))ds$ for $n\ge 0$. The condition $t\in [t_0-\alpha,t_0+\alpha]$ is to make sure that $x_{n+1}(t)$ does not fall out of the interval $[x_0-b,x_0+b]$ at any step $n$. If you know that $|f(t,x)|$ is at most $M$ and the interval $[t_0,t]$ is of length at most $\frac{b}{M}$ then the value of the integral $\int_{t_0}^t f(t,x_n(s))ds$ has modulus at most $\frac{b}{M} \cdot M = b$, so we are still in business with $x_{n+1}$.