Transversality of sections I am working through a paper https://arxiv.org/pdf/1011.1690.pdf on $J$-holomorphic surves and I don't understand the part in the theorem

where it says "smooth section transverse to the zero section". 
I could interpret this as $s(t) \neq 0_{section}(t)=0$ but this doesn't make sense to me..
Can somebody explain this to me?

First I will assume that $k\leq n$ so the statement makes more sense.
Let $z$ denote the zero section of $\pi \colon E\to B$. We have that $s$ and $z$ are transverse at $y\in s(B)\cap z(B)$ if for  $x = \pi(y)$
$$Ds_x(T_xB) + Dz_x(T_xB) = T_yE.$$
We say that the sections are transverse if they are transverse at every point of intersection of their images (which can be empty). 
Take a point $p\in B$ and a small neighborhood $V$ of $p$ in $B$ such that $E|_V \simeq V \times \mathbb{R}^k$ is trivial. Then in this trivialization we have 
$$
z\colon V \rightarrow V \times \mathbb{R}^k, \quad z(x) = (x,0) \\
s\colon V \rightarrow V \times \mathbb{R}^k, \quad s(x) = (x,f(x)) \\
$$
for some (smooth) function $f\colon V \rightarrow \mathbb{R}^k$. Now $s(p)= z(p)$  if and only if $f(p) =0$ and the transversality condition translates to $Df_p$ having maximal rank ($=k$). 
Note that $s^{-1}(0) = s^{-1}\left(s(B)\cap z(B) \right)$ is (locally) given by the system of equations $f(x)=0$ and the transversality condition allows us to apply the Implicit function theorem to prove that $s^{-1}(0)$ is a submanifold of dimension $n-k$. 
