I have a statement regarding transversal curves and their relation to the vector field that I cannot figure out why it is true. Not even at an intuitive level.
I quote,
"Since the vector field is transverse to the curve $\gamma$ the vectors $v(x)$ for all $x \in \gamma$ all point in the same direction. Else by continuity and mean-value theorem some $v(x)$ would be parallel to the transversal curve."
Why cant the curve just make a really steep turn after traveling a long way for instance, then come back in the other direction even in small neigbourhood?
I think there's a typo -- they mean "by continuity and intermediate-value theorem", not "by continuity and mean-value theorem".
The argument is that $v(\gamma(t))\times\gamma'(t)$ (the two-dimensional cross product) cannot be zero because $v(x)$ is nonzero, and $\gamma'(t)$ is (implicitly) supposed to be nonzero when saying the curve is smooth. These two nonzero orthogonal vectors cannot be parallel so their cross product is nonzero.
Since $v(\gamma(t))\times\gamma'(t)$ is also a continuous function of $t$ it cannot have values of different signs in different places without also being zero somewhere between them -- but we've just seen it is never zero.