Let $\mathbb R^n_a=\{(a,v)=v_a\mid a\in\mathbb R^n, v\in \mathbb R^n\}$ the set of vector of $\mathbb R^n$ with origin $a$. Let $$D_{v}|_af=D_vf(a)=\left.\frac{d}{dt}\right|_{t=0}f(a+tv)$$ where $D_v|_a:\mathcal C^\infty (\mathbb R^n)\to \mathbb R$. Let $T_a\mathbb R^n$ the set of all derivatives of $\mathcal C^\infty (\mathbb R^n)$ at $a$.
Why $v_a\mapsto D_v|_a$ is a linear map ?
Attempts
Let $f\in \mathcal C^\infty (\mathbb R^n)$. $$D_{v+w}|_af=\lim_{t\to 0}\frac{f(a+t(v+w))-f(a)}{t}$$ $$=\lim_{t\to 0}\frac{f(a+tv+tw)-f(a+tw)}{t}+\underbrace{\lim_{t\to 0}\frac{f(a+tw)-f(a)}{t}}_{=D_{w}|_a f},$$
Now, how can I prove that $$\lim_{t\to 0}\frac{f(a+tv+tw)-f(a+tw)}{t}=D_v|_af\ \ ?$$
Hint
If $v\in \mathbb R^n$, $$f(a+tv)=f(a)+t\nabla f(a)\cdot v+o(t).$$