If $b\in\beta$ and $b\not\in\alpha,$ then there is some $b'<b,$$b'>b,$ such that $b'\not\in\alpha,$$b'\in\beta,$ by the definition of a cut.
Let $r=b-b'>0.$$r=b'-b>0.$ (Intuitively, we have $\alpha+r\leq\beta.$$\alpha+r<\beta.$)
Let $g_1\in \gamma$ and $g_2\not\in\gamma$ such that $g_2-g_1<r.$
We see that $g_1+b\in \beta+\gamma.$$g_1+b'=g_1+b+r\in \beta+\gamma.$
If $g_1+b\in\alpha+\gamma,$$g_1+b+r\in\alpha+\gamma,$ then $$g_1+b=a+g_3\tag1$$$$g_1+b+r=a+g_3\tag1$$ for some $a\in\alpha,g_3\in\gamma.$
But $a<b'$$a<b$ and $g_3<g_2<g_1+r,$ so:
$$a+g_3<b'+g_1+r=b+g_1,$$$$a+g_3<b+g_1+r=b+g_1,$$ which contradicts $(1).$
We are essentially constructing $g_1,g_2$ as approximations of the real number $\gamma$ on either side of the cut so that $g_2-g_1$ is less than the intuitive real difference $\beta-\alpha.$
What we have is, intuitively, $g_1<\gamma<g_1+r$ and $\alpha <b'<b'+r<\beta$$\alpha <b<b+r<\beta.$
Then for any $a<\alpha,$ and $g<\gamma,$ $a< b'$$a< b$ and $g<g_1+r,$ so $$a+g<b'+g_1+r=g_1+(b'+r)<\beta+\gamma.$$$$a+g<b+g_1+r=g_1+(b+r)<\beta+\gamma.$$
This assumes you know that:
Given a cut $\gamma,$ and rational $r>0$ there are rationals $g,h$ with $g\in\gamma$ and $h\not\in\gamma,$ and $h-g<r.$
I suppose some of these theorems depend on how you define cuts which are rationals. I an assuming if $p$ is rational, the corresponding cut is $\pi_p=\{q\in\mathbb Q\mid q\leq p\}.$
But some books ensure that cuts do not contain their maximum, in which case what we need to be more specific than saying $b\not\in A.$