Two versions ...
Because $f(z)$ is entire, it has a Taylor expansion
$$f(z)=a_0+\sum\limits_{n=1}a_nz^n \tag{1}$$
$$f'(z)=a_1+\sum\limits_{n=2}na_nz^{n-1} \tag{2}$$
From
$$\left|f'(z)\right|\leq 2|z|, \forall z\in \mathbb{C} \Rightarrow |f'(0)|\leq 0$$
or
$$0=|f'(0)|=|a_1| \Rightarrow a_1=0 \Rightarrow f(z)=a_0+\sum\limits_{n=2}a_nz^n \tag{3}$$
But then
$$\left|f'(z)\right|\leq 2|z|, \forall z\ne0 \Rightarrow \left|\sum\limits_{n=2}na_nz^{n-1}\right|\leq 2|z| \Rightarrow |z|\left|\sum\limits_{n=2}na_nz^{n-2}\right|\leq 2|z|\Rightarrow\\
\left|\sum\limits_{n=2}na_nz^{n-2}\right|\leq 2, z\ne0$$
or
$$\left|\sum\limits_{n=2}na_nz^{n-2}\right|\leq \max\{2,2|a_2|\}, \forall z \in \mathbb{C}$$
This means that $g(z)=\sum\limits_{n=2}na_nz^{n-2}$, which is entire, is also bounded. According to Liouville's theorem $g(z)$ is constant. But $f'(z)=z\cdot g(z)=Cz$ or $f(z)=a_0+\frac{C}{2}z^2$ and the result follows ...
An alternative approach is to apply Cauchy's estimate to $(2)$
$$a_n=\frac{f^{(n)}(0)}{n!} \Rightarrow na_n=\frac{(f')^{(n-1)}(0)}{(n-1)!}=\frac{1}{2\pi}\int\limits_{C_R}\frac{f'(z)}{z^{n}}dz$$
leading to
$$|na_n|\leq \frac{1}{2\pi}\int\limits_{C_R}\left|\frac{f'(z)}{z^{n}}\right||dz|\leq \frac{1}{2\pi}\int\limits_{C_R}\left|\frac{2}{z^{n-1}}\right||dz|=\frac{2}{R^{n-2}}$$
Taking the $\lim\limits_{R\rightarrow\infty}$ we have $a_n=0,\forall n\geq 3$. As a result, considering $(3)$ too $$f(z)=\sum\limits_{n=1}a_nz^n=a_0+a_2z^2=a+bz^2$$
Last part, for $\forall z\ne 0$:
$$|f'(z)|\leq 2|z| \Rightarrow |2bz|\leq 2|z| \Rightarrow |bz|\leq |z| \Rightarrow |b|\leq 1$$
