Trouble understanding transversality I'm reading "Differential Topology" by Guillemin, V and Pollack, A. While reading the chapter about transversality, I got through this theorem :https://i.stack.imgur.com/5wYBo.jpg
(I'm not allowed to put pictures yet).
After the proof he says that this theorem easily implies that transverse maps are general when the target manifold ($Y$) in the euclidean space $\mathbb{R}^m$. If you take $S$ to be an open ball, $F(x,s)=f(x)+s$, and you fix $x\in X$ you get that $F$ is a traslation of the ball, and therefore a submersion. Everything ok until now. But then he says "So, of course, $F$ is a submersion of $X\times S$ and therefore transversal to any submanifold $Z\subset\mathbb{R}^m$". Can you help me with that step? Thank you very much.

It's an easy trick/observation. This sort of argument occurs many times in the text and in the exercises.
Because $S$ is an open ball in $\Bbb R^m$, in fact the $x$ variable is irrelevant. Fixing $x$, the derivative $dF_{(x,s)}$ maps $\{0\}\times T_sS \cong \Bbb R^m$ by the identity map to $\Bbb R^m$. That is, $dF_{(x,s)}(0,v)=v$ for all $v\in\Bbb R^m$. There's nothing more to do.
Any submersion $f\colon X\to Y$, by the way, is transverse to any submanifold $Z\subset Y$: For any $x\in f^{-1}(Z)$, $df_x\colon T_x X\to T_{f(x)}Y$ is surjective, so for any $Z$ whatsoever,
$$\text{im}(df_x)+ T_{f(x)}Z = T_{f(x)}Y.$$
