If $f'(x)\cdot x$ goes to zero then $f(2x)-f(x)$ is bounded.

Since $f'(x)\cdot x\to0$ as $\left\| x\right\|\to\infty$ then by choosing $\| x\|$ large enough (say for $\|x\|>R$) we can guarantee that $\|Df(x)\cdot x\|<1$. So for $x$ in this region we have that since $\mathbb{R}^{m}$ is convex we can apply mean value theorem and get:

$$\| g(x)\|=\| f(2x)-f(x)\|=\left\| \int_{0}^{1}(Df((1+t)x)\cdot x)\,dt\right\|$$

$$=\left\|\int_{0}^{1}\frac{1}{1+t}(Df((1+t)x)((1+t)x))\,dt\right\|\le\int_{0}^{1}\frac{1}{1+t}\,dt=\ln(2)$$

On the other hand $g$ is the difference of two differentiable functions hence on the complement of the previous region ($\| x\|\le R$) $g$ is bounded since this set is compact. A proof of the version of the mean value theorem used can be found here: http://en.wikipedia.org/wiki/Mean_value_theorem.

Here's a suggestion which is similar in spirit to your strategy, but with the fundamental theorem of calculus replacing the mean value theorem.

Let $\vec{v}$ be a unit vector in $\mathbb{R}^m$, and let $t>0$ be a scalar. Then $g(t\vec{v})$ can be computed with the fundamental theorem of calculus as $$g(t\vec{v}) = f(2t\vec{v}) - f(t\vec{v}) = \int_t^{2t}f'(s\vec{v})\cdot \vec{v}\,\,ds = \int_t^{2t}\frac{1}{s}\,[f'(s\vec{v})\cdot(s\vec{v})]\,\,ds.$$ What can you conclude?