A differentiable function on Euclidean Space compatible with scalar multiplication is a linear map

It's probably phrased in a confusing amount of 'generality'. All that we need is differentiability at $0$. Hence, let $Df(0)$ be the linear map given by the total derivative at $0$. All we need to argue is that $f(x)=Df(0)x$ for all $x$.

Let $x\in \mathbb{R}^n\setminus \{0\}$ and note that for all $\lambda \in \mathbb{R}\setminus \{0\}$

$$ f(x)=\lambda f\left(\frac{x}{\lambda}\right)=\lambda\left( Df(0)\frac{x}{\lambda}+o\left(\left\|\frac{x}{\lambda}\right\|\right)\right)=Df(0)x+\varepsilon\left(x/\lambda\right)\|x\|, $$ where $\varepsilon$ is some function with the property that $\lim_{\|y\|\to 0}\varepsilon(y)=0.$ However, the left-hand side is completely independent of $\lambda$, so we get that

$$ f(x)=Df(0)x+\lim_{\lambda\to \infty}\varepsilon(x/\lambda)\|x\|=Df(0)x $$

First observe that $f(0)=0$. Now fix $x,y\in \Bbb{R}^n$. For every positive $\lambda$ we know: $$ f(x+y) -f(x)-f(y)=\frac{\lambda(f(x+y) -f(x)-f(y))}{\lambda}=\frac{f(\lambda(x+y)) -f(\lambda x)-f(\lambda y)}{\lambda}$$ This shows the function $\lambda \to\frac{f(\lambda(x+y)) -f(\lambda x)-f(\lambda y)}{\lambda}$ defined for positive $\lambda$ is constant. What about its limit as $\lambda \to 0$?