Consider a map $T:X\to Y$ of the form $$T(u):=p(u)H(u),$$ where $H:X\to Y$ is a Lipschitz homogeneous map (here always meaning $H(\alpha u)=\alpha H(u)$ for all $u\in X$ and $\alpha\in\mathbb{K}$), and $p:X\to\mathbb{R}$ is a bounded function with the property that the image of any ball of fixed radius $R$ is the same. For instance, any bounded $R$-periodic function of the norm.
So for any $u\in X$ and $\alpha\in\mathbb{K}$ there is $v\in X$ such that $\| \alpha u -v\|\le R$ and $p(v)=p(u)$. Then $\alpha T(u)-T(v)=\alpha p(u) H(u)-p(v)H(v)=p(u)(H(\alpha u) -H(v)) $ so that $\|\alpha T(u)-T(v)\|\le\|p\|_\infty\operatorname{Lip}(H)R$. Thus $T$ satisfies your hypothesis with $N:=\max\big(1, \|p\|_\infty\operatorname{Lip}(H)\big)R.$
On the other hand, if a map $T$ satisfies your hypothesis, and also has the following mild property: for any $u\in X$
$$\sup_{\|\alpha u-w\|\le N} \|T(w)-T(\alpha u)\|=o(\alpha)\qquad (\text{for}\; |\alpha|\to+\infty),$$
(for instance, it is Lipschitz),
then it is a homogeneous map. For, if $v$ is as in your notation,
$$\alpha T(u)= T(\alpha u)+ \big( T(v)-T(\alpha u)\big)+ \big(\alpha T(u)-T(v)\big)= T(\alpha u)+o(\alpha)$$
so that $T(0)=0$, and for all $u\in X$ $$T(u)=\lim_{|\alpha|\to +\infty}{1\over\alpha}T(\alpha u)$$
which implies it is a homogeneous map.