Dynamic system in the book (chapter 6, page 67 in
http://www.me.berkeley.edu/ME237/6_cont_obs.pdf)
$$ \begin{cases} \dot{x}_{1}=x_{2}^{2}\\ \dot{x}_{2}=u\end{cases} $$ so
$$ f=\begin{bmatrix} x_{2}^{2} \\ 0 \\ \end{bmatrix}, g=\begin{bmatrix} 0 \\ 1 \\ \end{bmatrix}, $$
then
$$ C=[g,[f,g]]= \begin{bmatrix} 0 & -2x_2\\ 1 & 0\\ \end{bmatrix} $$
where $[f,g]$ denotes the Lie Bracket. Authors conclude system is not controllable because on $x_{2}=0$ $C$ loses rank. This is straigforward to prove since $\dot{x_1}>0$ so "I can't go back".
But let now consider
$$ \begin{cases} \dot{x}_{1}=s(x_{2})=\frac{1}{1+e^{-x_{2}}}\\ \dot{x}_{2}=u\end{cases} $$
where $s(x)$ is the well known sigmoid function applying the same analysis we reach
$$ C=[g,[f,g]]= \begin{bmatrix} 0 & -s(x_{2})(1-s(x_{2}))\\ 1 & 0\\ \end{bmatrix} $$
so C never loses rank, so it would be saying system is controllable. However always $\dot{x}_{1}>0$ so "I can't go back" just like in previous system so it's not controllable.
Am I missing something?
In a non-linear system you should talk first about local accessibility, and from this point you can try to extend it to global or semi-global.
Note that C loosing his rank is a sufficient condition but not a necessary one. In your notes the authors says "if" and not "if and only if". Hence, there is not a contradiction in your question, you can have a "can not go back" and C being always full rank at the same time.