I want to show that
$$ D_{v}f(p)=\sum_{i}^{n}\frac{\mbox{d}c^{i}}{\mbox{dt}}(0)\cdot\frac{\partial f}{\partial x_{i}}(p) $$
My proof:
We define $c(t)=p+tv$ so that we can write $$ D_{v}f(p)=\frac{\mbox{d}}{\mbox{dt}}\Big|_{t=0}(f\circ c)(t)=D_{1}(f\circ c)(0) $$
We now apply the chain rule and get
$$ D_{1}(f\circ c)(0)=D(f\circ c)(0)(1)=\left(Df\left(c(0)\right)\circ Dc(0)\right)(1)= $$
$$ =Df\left(p\right)\left(D_{1}c(0)\right)=Df(p)(\frac{\mbox{d}}{\mbox{dt}}\Big|_{t=0}c(t))=Df(p)(v) $$
(notice that we could have just used the fact that $D_{v}f(p)=Df(p)(v)$, but this way we re-derived this result using the chain rule).
Now, $$ Df(p)(v)=Df(p)(\sum_{i}^{n}v^{i}e_{i})=\sum_{i}^{n}v^{i}Df(p)(e_{i})=\sum_{i}^{n}\frac{\mbox{d}c^{i}}{\mbox{dt}}(0)\cdot\frac{\partial f}{\partial x_{i}}(p) $$
where the last equality holds because $v^{i}=\left(\frac{\mbox{d}c}{\mbox{dt}}(0)\right)^{i}=\frac{\mbox{d}c^{i}}{\mbox{dt}}(0)$.
My question:
I used $v^{i}=\frac{\mbox{d}c^{i}}{\mbox{dt}}(0)$ but it seems to me that this would be true for any $t_0$: $v^{i}=\frac{\mbox{d}c^{i}}{\mbox{dt}}(t_0)$. Is this correct?
To answer your question, yes because $v$ is a constant vector. In reference to the flow of the argument, this way below is much shorter.
\begin{align*} D_v f(p) = \frac{d}{dt} \Bigr|_{t = 0} f(p+tv) &= \nabla f(p) \cdot v \\& = \sum_j \frac{\partial f}{\partial x^j}(p) \cdot v^j \\ & = \sum_j \frac{\partial f}{\partial x^j} \cdot \frac{d c^j}{dt}\Bigr|_{t = 0} \end{align*}
The last equality follows from the fact that,
$$c(t) = p+tv = (p_1 + t v_1,...,p_n + t v_n) = \begin{pmatrix} c^1 \\ \vdots \\ c^n \end{pmatrix}$$