I'm using the $3rd$ edition of the book. I'm having trouble seeing the follow two estimates in the proof of Theorem 6.2 from pages 91 and 92.
For the first estimate I'm having trouble with is right afterthe sentence "On the other hand, if $|x_o-y_o|\geq d/2$," $$d_{x_o}^{2+\alpha}\frac{|D^2u(x_o)-D^2u(y_o)|}{|x_o-y_o|^\alpha}\leq \left(\frac{2}{\mu}\right)^\alpha[d_{x_o}^2|D^2u(x_o)|+d^2_{y_o}|D^2u(y_o)|].$$ Originally, I though this followed from the chain $$(2^2 |x_o-y_o|)^\alpha \geq (2d)^\alpha \geq(d_{x_o})^\alpha,$$ but this can't be since $d_{x_o}/2\geq d$, not the other way around.
The second question regards the first inequality of (6.18) on page 92, $$|g|_{0,\alpha;B}^{(2)}\leq d^2|g|_{0;B}+d^{2,\alpha}[g]_{\alpha;B}.$$
First question: By the triangle inequality, $$ d_{x_0}^{2+\alpha} \frac{\vert D^2u(x_0)-D^2u(y_0)\vert}{\vert x_0-y_0\vert^\alpha} \leqslant \bigg ( \frac{d_{x_0}}{\vert x_0-y_0\vert}\bigg )^\alpha \big ( d_{x_0}^2\vert D^2u(x_0)\vert + d_{x_0}^2 \vert D^2u(y_0)\vert \big ). $$ On one hand, they assume that $d_{x_0}=d_{x_0,y_0}=\min \{ d_{x_0},d_{y_0}\}$, so $d_{x_0} \leqslant d_{y_0}$. Hence, $$ d_{x_0}^2\vert D^2u(x_0)\vert + d_{x_0}^2 \vert D^2u(y_0)\vert \leqslant d_{x_0}^2\vert D^2u(x_0)\vert + d_{y_0}^2 \vert D^2u(y_0)\vert. $$ On the other hand, since $\vert x_0-y_0\vert \geqslant d/2$ and $d=\mu d_{x_0}$, we have that $$\bigg ( \frac{d_{x_0}}{\vert x_0-y_0\vert}\bigg )^\alpha \leqslant \bigg ( \frac{2d_{x_0}}d\bigg )^\alpha = \bigg ( \frac{2}\mu\bigg )^\alpha .$$ These two observations give you that $$\frac{\vert D^2u(x_0)-D^2u(y_0)\vert}{\vert x_0-y_0\vert^\alpha} \leqslant \bigg ( \frac{2}\mu\bigg )^\alpha \big (d_{x_0}^2\vert D^2u(x_0)\vert + d_{y_0}^2 \vert D^2u(y_0)\vert \big ) $$ as required (I believe that the $\alpha$ on the denominator of the RHS of your question is a typo and should be $\mu$.)
Second question: By definition, $$\vert g \vert^{(2)}_{0,\alpha;B}= \vert g \vert^{(2)}_{0;B}+[ g ]^{(2)}_{0,\alpha;B}.$$ Then $$\vert g \vert^{(2)}_{0;B} =\sup_{x\in B} d_{\partial B}^2(x)\vert g(x)\vert$$ where $d_{\partial B}(x) = \operatorname{dist}(x,\partial B)$. Since $B=B_d(x_0)$, we have that $d_{\partial B}(x) \leqslant d$, so $$ \vert g \vert^{(2)}_{0;B} \leqslant d^2\sup_{x\in B} \vert g(x)\vert= d^2\vert g\vert_{0;B}.$$ Similarly, $$[ g ]^{(2)}_{0,\alpha;B} =\sup_{\substack{x,y\in B \\ x\neq y}} \bigg \{ d_{x,y}^{2+\alpha} \frac{\vert g(x)-g(y)\vert}{\vert x-y\vert^\alpha}\bigg \} \leqslant d^{2+\alpha} \sup_{\substack{x,y\in B \\ x\neq y}} \bigg \{ \frac{\vert g(x)-g(y)\vert}{\vert x-y\vert^\alpha}\bigg \} =d^{2+\alpha} [g]_{0,\alpha;B}$$