If $f(x)$ and $(f(2x)-f(x))/x$ have limit $0$ as $x\to 0$, then $f(x)/x\to 0$

As pointed out by Ilya the argument of the OP is incorrect. Here is my proposal:

There is a function $x\to g(x)$ with $g(x)\to 0$ $\ (x\to 0)$ such that

$$f(2x)-f(x)=x\ g(x)\ .$$

As $f(x)\to 0$ $\ (x\to 0)$ we can set up the following telescopic series:

$$f(x)\ =\ \sum_{k=1}^\infty \bigl(f(2\cdot 2^{-k}x)- f(2^{-k}x)\bigr) \ =\ x\ \sum_{k=1}^\infty 2^{-k} g(2^{-k}x)\ .$$

Since $\sum_{k\geq 1} 2^{-k}=1$ it follows that

$$\left|{f(x)\over x}\right|\ \leq\ \sup_{0<|t|<|x|}\ |g(t)|\to 0\qquad (x\to 0)\ .$$

This is Problem 3.2.7 from Problems in real analysis by Teodora-Liliana T. Rădulescu, Vincentiu D. Radulescu, Titu Andreescu; p.121-122. The only difference is that they use $\lim\limits_{x\to 0} \frac{f(x)-f(x/2)}x=0$, which is obviously equivalent to the condition from the question, and they work with $x>0$, which can be easily modified. I'll give a sketch of their solution here.

The basic idea is to notice $\lim\limits_{x\to 0} \frac{f(x)-f(x/2)}x=0$ implies that for given $\varepsilon>0$ there is $\delta>0$ such that $$|f(x)-f(x/2)|<\varepsilon |x|$$ whenever $x<\delta$. For any fixed $x<\delta$ we have $|f(x/2^n)-f(x/2^{n+1})|<\varepsilon |x|/2^n$ and by triangle inequality we get $$|f(x)-f(x/2^n)| \le 2\varepsilon |x|.$$ Now taking the limit $n\to\infty$ and using $\lim\limits_{x\to 0} f(x)=0$ gives $$|f(x)|\le 2\varepsilon |x|$$ $$\frac{|f(x)|}{|x|} \le 2\varepsilon$$ for any $x<\delta$.