confusion about chain rule in linearity proof Suppose $v,p\in\mathbb{R}^n$, then $D_v|_p:C^\infty\to\mathbb{R}$ is defined by $$D_v|_p(f):=\frac{d}{dt}|_{t=0}f(p+tv).$$
Now I'm looking at a proposition that states that the map $v\mapsto D_v|_p$ is a linear isomorphism from $\mathbb{R}^n\to T_p\mathbb{R}^n$. I understand most of the proof, but I'm confused by the linearity part:
$$\frac{d}{dt}|_{t=0}f(p+tv+\lambda tw) = \frac{d}{dt}|_{t=0}f(p+tv) + \lambda\cdot\frac{d}{dt}|_{t=0}f(p+tw)$$
for $v,w,p\in\mathbb{R}^n$ and $\lambda\in\mathbb{R}$.  
This equality should immediately follow from the chain rule, but I don't see how. Probably I'm just a bit confused by the notation of (directional) derivatives... but I already searched quite a bit and just can't see how the chain rule is used here. Can someone explain this?

We have
$$
D_v|_p(f):=\left.\dfrac{d}{dt}\right|_{t=0}f(p+tv)
$$
Set $h(t)=p+tv.$ Then
$$
D_p(f)(v):=\left.\dfrac{d}{dt}\right|_{t=0}f(p+tv)=\left.\dfrac{d}{dt}\right|_{t=0}f(h(t))=f'(h(0))\cdot h'(0)=f'(p)\cdot v 
$$
Next, set $g(t)=p+tv+\lambda tw.$ It is $f'=D_p(f)\cdot v$ reading "Derivative at point $x=p$ of function $f$ (= gradient = Jacobi matrix = linear approximation) in direction of $v$."
\begin{align*}
D_p(f)\cdot (v+\lambda w)&=
\left. \frac{d}{dt}\right|_{t=0}f(p+t(v+\lambda w)) \\&=\left. \frac{d}{dt}\right|_{t=0}f(g(t)) \\
&=\left. \frac{d}{dt}\right|_{g(t)=g(0)}f((g(t))  \cdot \left. \frac{d}{dt}\right|_{t=0}g(t)\\
&=\left. \frac{d}{ds}\right|_{s=0}f(s)\cdot (v+\lambda w)\\
&=f'(p)\cdot v +\lambda \cdot f'(p)\cdot w\\
&=\left. \dfrac{d}{dt}\right|_{t=0}f(p+tv)+\lambda \cdot \left. \dfrac{d}{dt}\right|_{t=0}f(p+tw)\\
&=D_p(f)\cdot v +\lambda \cdot D_p(f)\cdot w\\
&=D_p(f)(v)+\lambda D_p(f)(w)
\end{align*}
