How $b>a \lor a \leq 0 \implies \max⁡(b,0)-\max⁡(\min⁡(a,b),0) \geq 0$?

You can separate the cases.

Case 1: $a\leq 0$.

If $a\leq 0$, then you know that $\min(a,b)\leq 0$, which means that $\max(\min(a,b),0)=0$.

You also know that $\max(b,0)\geq 0$, which means that $$\max(b,0)+\max(\min(a,b),0) = \max(b,0)+0=\max(b,0)\geq 0.$$

Case 2: $a>0$.

In this case, because we know that $b>a\lor a\leq 0$, we know that $b>a$.

Therefore, we know that $\min(a,b)=a$, which means that $\max(\min(a,b),0) = \max(a,0)=a$, because $a>0$.

This means that

$$\max(b,0)-\max(\min(a,b),0)=\max(b,0)-a=b-a,$$ and since we know $b>a$, we also know $b-a>0$, which means the entire expression is positive in this case.

If you want to fully write out the range of values that the expression $\max(b,0)-\max(\min(a,b),0)$ can take, then you need to separate out some more cases. In particular, you have the following $6$ cases:

$$\max(b,0)-\max(\min(a,b),0)=\begin{cases}0&\text{if}&a\leq b\leq 0\\ b&\text{if}&a\leq 0\leq b\\ 0&\text{if}&b\leq a \leq 0\\ 0&\text{if}&b\leq 0\leq a\\ b-a&\text{if}&0\leq a\leq b\\ 0&\text{if}&0\leq b\leq a\end{cases}$$

This also shows that the expression is always positive (i.e., not only when $b<a\lor a\leq 0$).