$$w\left(a,b\right)=a-\sqrt{a^2-b}$$
$$w\left(\alpha ,\beta \right)=\alpha -\sqrt{\alpha ^2-\beta }=a\left(1+E_a\right)-\sqrt{\left(a\left(1+E_a\right)\right)^2-b\left(1+E_b\right)}$$
$$\frac{\left|w\left(\alpha ,\beta \right)-w\left(a,b\right)\right|}{\left|w\left(a,b\right)\right|}$$
Let's look what's going on inside the numerator:
$$w\left(\alpha ,\beta \right)-w\left(a,b\right) = a\left(1+E_a\right)-\sqrt{\left(a\left(1+E_a\right)\right)^2-b\left(1+E_b\right)}-\left(a-\sqrt{a^2-b}\right)$$
$$=a+aE_a-\sqrt{a^2+2E_aa^2+a^2E_a^2-b-bE_b}-a+\sqrt{a^2-b}=aE_a-\sqrt{a^2+2E_aa^2+a^2E_a^2-b-bE_b}+\sqrt{a^2-b}$$
Thus:
$$\left|w\left(\alpha ,b\right)-w\left(a,b\right)\right|=\left|aE_a-\sqrt{a^2+2E_aa^2+a^2E_a^2-b-bE_b}+\sqrt{a^2-b}\right|\le \left|aE_a\right|+\left|\sqrt{a^2+2E_{a}a^2+a^2E_a^2-b-bE_b}\right|+\left|\sqrt{a^2-b}\right|$$
We know that $\sqrt{x+y}\le \sqrt{x}+\sqrt{y}$, thus (if $x=a^2-b,\:y=2E_aa^2+a^2E_a^2-bE_b$):
$$\left|aE_a\right|+\left|\sqrt{a^2+2E_a^2+a^2E_aa^2-b-bE_b}\right|+\left|\sqrt{a^2-b}\right|\le \left|aE_a\right|+\left|\sqrt{a^2-b}\right|+\left|\sqrt{2E_aa^2+a^2E_a^2-bE_b}\right|+\left|\sqrt{a^2-b}\right|$$
$$=\left|aE_a\right|+2\left|\sqrt{a^2-b}\right|+\left|\sqrt{2E_aa^2+a^2E_a^2-bE_b}\right|$$
We also know that:
$$\left|aE_a\right|+2\left|\sqrt{a^2-b}\right|+\left|\sqrt{2E_aa^2+a^2E_a^2-bE_b}\right|\le \left|aE_a\right|+\left|\sqrt{2E_aa^2+a^2E_a^2-bE_b}\right|$$
We use that $\left|E_{a,b}\right|\le 2^{-t}$
$$\left|aE_a\right|+\left|\sqrt{2E_aa^2+a^2E_a^2-bE_b}\right|\le \left|a\right|2^{-t}+\left|\sqrt{2^{-t}\left(2a^2-b\right)+2^{-2t}\left(a^2\right)}\right|$$
We split the radix similary as before:
$$\left|a\right|2^{-t}+\left|\sqrt{2^{-t}\left(2a^2-b\right)+2^{-2t}\left(a^2\right)}\right|\le \left|a\right|2^{-t}+\left|\sqrt{2^{-t}\left(2a^2-b\right)}+\sqrt{2^{-2t}\left(a^2\right)}\right|\le \:\left|a\right|2^{-t}+\left|\sqrt{2^{-t}\left(2a^2-b\right)}\right|+\left|\sqrt{2^{-2t}\left(a^2\right)}\right|=\left|a\right|2^{-t}+2^{-\frac{1}{2}t}\left|\sqrt{2a^2-b}\right|+\left|a\right|2^{-t}=2\left|a\right|2^{-t}+2^{-\frac{1}{2}t}\left|\sqrt{2a^2-b}\right|$$
We use that:
$$2\left|a\right|2^{-t}+2^{-\frac{1}{2}t}\left|\sqrt{2a^2-b}\right|\le 2\left|a\right|2^{-t}+2^{-t}\left|\sqrt{2a^2-b}\right|=2^{-t}\left(2\left|a\right|+\left|\sqrt{2a^2-b}\right|\right)$$
Thus in the end, we have:
$$\frac{\left|w\left(\alpha \:,\beta \:\right)-w\left(a,b\right)\right|}{\left|w\left(a,b\right)\right|}\le 2^{-t}\frac{\left(2\left|a\right|+\left|\sqrt{2a^2-b}\right|\right)}{\left|a-\sqrt{a^2-b}\right|}$$
So that $\frac{\left(2\left|a\right|+\left|\sqrt{2a^2-b}\right|\right)}{\left|a-\sqrt{a^2-b}\right|}$ is our condition number
Am I right?