The question is a bit of a mouthful. In Classical Mechanics by Goldstein I have seen the use of the following:
$$ \frac{d\dot{F}}{d\dot{q}_i} = \frac{dF}{dq_i} $$
where $F = F(q_1,q_2,...,q_n;t)$, $ \dot{F} = \frac{dF}{dt} $ and $ \dot{q_i} = \frac{dq_i}{dt} $ . How would I show this?
The use of this result is used on page 17 of the PDF solutions guide (Page 10 within the derivations chapter) by showing that the transformation $L'(q,\dot{q},t) = L(q,\dot{q},t) + \frac{dF(q,t)}{dt}$ keeps the form of the Euler-Lagrange equation invariant:
http://www.slideshare.net/venuatsrr/solution-manual-classical-mechanics-goldstein
In general, this is not true. For instance, suppose the curve is parametrized by $x=t^2$, $y=t$. Then $$ \frac{dy}{dx} = \frac{\dot y}{\dot x} = \frac{1}{2t} = \frac{1}{2y} $$ But $\dot y$ is constant, so $$ \frac{d\dot y}{d\dot x} = 0 $$
I looked at your reference, and as far as I can tell, your question is about this statement: If $F = F(q_1,q_2,\dots,q_n;t)$, then $$ \frac{\partial \dot F}{\partial \dot q} = \frac{\partial F}{\partial q} $$ The reason for this is the chain rule.
For each $i$, the expression $\dfrac{\partial \dot F}{\partial \dot q_i}$ stands for $\dfrac{\partial}{\partial \dot q_i}\dfrac{dF}{dt}$. By the chain rule: \begin{align*} \frac{dF}{dt} &= \frac{\partial F}{\partial q_1}\frac{dq_1}{dt}+\dots +\frac{\partial F}{\partial q_n}\frac{dq_n}{dt} +\frac{\partial F}{\partial t} \\ &= \frac{\partial F}{\partial q_1}\dot q_1+\dots +\frac{\partial F}{\partial q_n}\dot q_n +\frac{\partial F}{\partial t} \\ \end{align*} Then take $\dfrac{\partial}{\partial \dot q_i}$ of each side.