Condition (2) implies (1) with $5|y|$ instead of $2|y|$. Fix $y$, let $r=|y|$ and $I=\int_{|x| >5|y| }|K(x-y)-K(x)|\, dx$. Then $$I \leq \int_{|x| >5r }|K(x-y)-K(x-z)|\, dx+\int_{|x| >5r }|K(x-z)-K(x)|\, dx:=I_1(z)+I_2(z) $$ for every $|z| \leq r$. If $K$ is the supremum in (2), then $r^{-n} \int_{B(0,r)} I_2(z)\, dz \leq K$. In $I_1$ we set $\xi=x-z$ so that $|\xi| \geq 4r$ and $$I_1(z) \leq \int_{|\xi| \geq 4r} |K(\xi-(y-z))-K(\xi)|\, d\xi.$$ Since $|y-z| \leq 2r$, then $$r^{-n}\int_{B(0,r)} I_1(z)\, dz \leq r^{-n} \int_{B(0,2r)}|K(\xi-w)-K(\xi)|\, dw \leq 2^n K.$$ The estimate of $I$ in terms of $K$ now follows by averaging the inequality $I \leq I_1(z)+I_2(z)$ over $B(0,r)$.

Condition \eqref{2} implies \eqref{1} with $5|y|$ instead of $2|y|$. Fix $y$, let $r=|y|$ and $I=\int_{|x| >5|y| }|K(x-y)-K(x)|\, dx$. Then $$I \leq \int_{|x| >5r }|K(x-y)-K(x-z)|\, dx+\int_{|x| >5r }|K(x-z)-K(x)|\, dx:=I_1(z)+I_2(z) $$ for every $|z| \leq r$. If $K$ is the supremum in \eqref{2}, then $r^{-n} \int_{B(0,r)} I_2(z)\, dz \leq K$. In $I_1$ we set $\xi=x-z$ so that $|\xi| \geq 4r$ and $$I_1(z) \leq \int_{|\xi| \geq 4r} |K(\xi-(y-z))-K(\xi)|\, d\xi.$$ Since $|y-z| \leq 2r$, then $$r^{-n}\int_{B(0,r)} I_1(z)\, dz \leq r^{-n} \int_{B(0,2r)}|K(\xi-w)-K(\xi)|\, dw \leq 2^n K.$$ The estimate of $I$ in terms of $K$ now follows by averaging the inequality $I \leq I_1(z)+I_2(z)$ over $B(0,r)$.