From Taylor's Theorem we have $f(x)=f(0)+f'(0)x+\frac{f''(0)}{2}x^2+h(x)x^2=f(0)+x+\frac{1}{2}x^2+h(x)x^2, \ x \in \mathbb{R},$ where $h$ is a function s.t. $\displaystyle{\lim_{x \to 0}h(x)=0}.$ The polynomial $f(0)+x+\frac{1}{2}x^2$ is unique with the above property.
Since $g(x)=f\left(x^{10}\right)$ and $f$ is $13$ times differentiable $\Rightarrow \ \ g$ is $13$ times differentiable. Now $g(x)=f\left(x^{10}\right)=f(0)+x^{10}+\frac{1}{2}x^{20}+h\left(x^{10}\right)x^{20}=\\ f(0)+x^{10}+x^{13}\left(\frac{1}{2}x^{7}+h\left(x^{10}\right)x^{7}\right), \ x \in \mathbb{R}.$
Since $\displaystyle{\lim_{x \to 0}\left(\frac{1}{2}x^{7}+h\left(x^{10}\right)x^7\right) =0}$ from uniqueness in Taylor's theorem we conclude that $f(0)+x^{10}=g(0)+g'(0)x+\frac{g''(0)}{2}x^2+\ldots +\frac{g^{(13)}(0)}{13!}x^{13}$.
Therefore $g^{(11)}(0)=0.$
Actually $g^{(k)}(0)=0, \ \ \ \forall k \in \{1,2,\ldots,9,11,12,13\}$ and
$g(0)=f(0), \ g^{(10)}(0)=10!.$
The fact that $f^{(12)}$ is differentiable is important because to apply Taylor's theorem to $g$ we need to know how many times $g$ is differentiable. Of course he could have said that $f^{(10)}$ is differentiable.