Let $K:\mathbb R^n\backslash \{0\}\longrightarrow \mathbb C$ differentiable s.t. $|K(x)|\leq B|x|^{-n}$, and $|\nabla K(x)|\leq B|x|^{-n-1}$. Suppose $|x|>2|y|$. By IVT, $$K(x-y)-K(x)=-\int_0^1 \nabla _x K(x-ty)\cdot yd t.$$ Then, $$|K(x-y)-K(x)|\leq \int_0^1|\nabla _x K(x-ty)||y|d y\leq B|y|\int_0^1 |x-ty|^{-n-1}d t.$$
Q1) But I don't see why $$\int_0^1 |x-ty|^{-n-1}d t\leq \left(\frac{|x|}{2}\right)^{-n-1}.$$
Suppose that we have when $|x|>2|y|$ that $$|K(x-y)-K(x)|\leq 2^{n+1}B|y||x|^{-n-1}.$$ So $$\int_{|x|>2|y|}|K(x)-K(x-y)|dx\leq 2^{n+1}B|y|\int_{|x|>2|y|}|x|^{-n-1}dx\leq B_1 |y||y|^{-1}=B_1.$$
Q2) I don't understand the inequality $$2^{n+1}B|y|\int_{|x|>2|y|}|x|^{-n-1}dx\leq B_1 |y||y|^{-1}.$$ I would think that $$|x|>2|y|\implies \frac{1}{|x|}<\frac{1}{2|y|}\implies |x|^{-n-1}<2^{-n-1}|y|^{-n-1}$$ and thus $$2^{n+1}B|y|\int_{|x|>2|y|}|x|^{-n-1}dx\leq B|y||y|^{-n-1}\int_{|x|>2|y|}dx=+\infty ,$$
and thus it doesn't work...
For question $1$, since $|x|>2|y|$, we then have $$|x-ty|\ge |x|-t|y| \ge |x|-\frac{t|x|}{2} \ge \frac{|x|}{2}$$ since $t\in(0,1)$, and hence $$|x-ty|^{-n-1}\le\left(\frac{|x|}{2}\right)^{-n-1}.$$ For question 2, evaluate the integral directly using polar coordinates: $$\int\limits_{|x|>2|y|}{|x|^{-n-1}\text{ d}x} = \int\limits_{2|y|}^{\infty}{r^{-n-1}(B_r\text{ d}r)} = \int\limits_{2|y|}^{\infty}{r^{-n-1}B_1r^{n-1}\text{ d}r} = B_1\int\limits_{2|y|}^{\infty}{r^{-2}\text{ d}r} = \frac{B_1}{2|y|} $$ where $B_r$ is the volume of the $(n-1)$-sphere of radius $r$ (and hence $B_r = B_1r^{n-1}$).