We have :
$$
\|v\|^2_{L^2(\Omega)} = \left|\int_{\Gamma} \frac{x}{d}\cdot \eta v^2 ds - \int_{\Omega} \frac{|x|}{d} (2v \nabla v) ds \right| \\ \leq \max_{\Omega_1}\frac{|x|}{d}\left|\int_{\Gamma} v^2 ds\right| + \max_{\Omega_1} \frac{2|x|}{d} \left|\int_{\Omega} v \nabla v ds\right| \\ \leq \max_{\Omega_1} \frac{|x|}{d}\|v\|^2_{L^2(\Gamma)} + \max_{\Omega_1} \frac{2|x|}{d} \left|\int_{\Omega} v^2 ds\right|^{\frac 12} \left|\int_{\Gamma} |\nabla v|^2ds\right|^{\frac 12}
$$
by Cauchy-Schwarz. We trudge on :
$$
\max_{\Omega_1} \frac{|x|}{d}\|v\|^2_{L^2(\Gamma)} + \max_{\Omega_1} \frac{2|x|}{d} \left|\int_{\Omega} v^2 ds\right|^{\frac 12} \left|\int_{\Gamma} |\nabla v|^2ds\right|^{\frac 12}\\ = \max_{\Omega_1} \frac{|x|}{d}\|v\|^2_{L^2(\Gamma)} + \max_{\Omega_1} \frac{2|x|}{d} \|v\|_{L^2(\Omega)} \|\nabla v\|_{L^2(\Omega)}
$$
whence we are at :
$$
\|v\|^2_{L^2(\Omega)}\leq \max_{\Omega_1} \frac{|x|}{d}\|v\|^2_{L^2(\Gamma)} + \max_{\Omega_1} \frac{2|x|}{d} \|v\|_{L^2(\Omega)} \|\nabla v\|_{L^2(\Omega)}
\\ \leq C'(\|v\|^2_{L^2(\Gamma)} + \|v\|_{L^2(\Omega)} \|\nabla v\|_{L^2(\Omega)})$$
Now, desperate times call for beautiful inequalities. Infact, the entirety of PDE theory is littered with inequalities that will blow anyone's mind, from the sublime to the ridiculous.
The inequality we use is this one. Recall that for any real $a,b$ we have $a^2+b^2 \geq 2ab$. We use this to write for any $C>0$ :
$$
2ab = 2\left(\frac aC\right) (bC) \leq \frac{a^2}{C^2} + C^2b^2 \implies ab \leq \frac{a^2}{2C^2} + 2C^2b^2
$$
for any real numbers $a$ and $b$. Let $a = \|v\|_{L^2(\Omega)} , b = \|\nabla v\|_{L^2(\Omega)}$. Then we have :
$$
\|v\|_{L^2(\Omega)} \|\nabla v\|_{L^2(\Omega)} \leq \frac{\|v\|^2_{L^2(\Omega)}}{2C^2} + 2 \|\nabla v\|^2_{L^2(\Omega)} C^2
$$
Put this back into the expression, and we get :
$$
\|v\|^2_{L^2(\Omega)} \leq C'(\|v\|^2_{L^2(\Gamma)} + \|v\|_{L^2(\Omega)} \|\nabla v\|_{L^2(\Omega)}) \\ \leq C'\left(\|v\|^2_{L^2(\Gamma)} + \frac{\|v\|^2_{L^2(\Omega)}}{2C^2} + 2 \|\nabla v\|^2_{L^2(\Omega)} C^2\right) \\
\leq C'\|v\|^2_{L^2(\Gamma)} + 2C'C^2 \|\nabla v\|^2_{L^2(\Omega)} + \frac{C'}{2C^2}\|v\|^2_{L^2(\Omega)}
$$
since we can choose $C$ as anything we want, we ensure that it is large enough so that $2C^2 > C'$. Then we can take $\frac{C'}{2C^2}\|v\|^2_{L^2(\Omega)}$ to the other side of the inequality, factorize and bring the $1 - \frac{C'}{2C^2}$ to the other side to get :
$$
\|v\|^2_{L^2(\Omega)} \leq \frac{C'}{1 - \frac{C'}{2C^2}} \|v\|^2_{L^2(\Gamma)} + \frac{2C'C^2}{1 - \frac{C'}{2C^2}} \|\nabla v\|^2_{L^2(\Omega)} \\ \leq D\left( \|\nabla v\|^2_{L^2(\Omega)} + \|v\|^2_{L^2(\Gamma)}\right)
$$
where $D$ is the larger of the coefficients. Hence, we are done!
The inequality I used is often seen in interpolation-type inequalities for Sobolev spaces. You will do well to remember it.
In general, to see the tricks and traps that PDE theory uses, it is recommended that you hike your way through the proofs of difficult results in the field, such as the Sobolev inequality , the embedding theorems, the Rellich-Kondrasov inequality, and the Poincare inequality (of course). Each of the proofs contain key algebraic observations like the one I make above, that allow you to go from one end to the other smoothly. These may be found in Kesavan's book titled "Topics in functional analysis and applications".
