To define $\chi(M)$ for a manifold with boundary first we need to show the existence of an outward-pointing vector field that is transverse to the zero section , at least as Hirsch does it. Now I am trying to understand why said vector field exists. It's clear to me that outward-pointing sections exists, and by transversality we have that the set of maps from $M$ to $TM$ that are transversal to the zero section is open and dense. Now I don't see how I can put these two things together to get the existence of a section that is outward-pointing and transversal to the zero section. I know I can find a section that is transverse to the zero section but this doesn't have to be outward-pointing.
Now in something similar that came up in my mind. Suppose we have a submanifold $N$ of $M$ of dimension $n-1$, is it possible to have a section that is tangent to $N$ at points of $N$ , where we make the canonical identifications with the inclusion map, such that this vector field is also transverse to the zero section of $M$? I thought about using Parametric transversality to the normal bundle and to the zero section but that got me nowhere. I know I can build a vector field such that when restricted to $N$ is tangent to it but when I try to make it transversal to the zero section by approximation I could lose this property.
Any help is appreciated. Thanks in advance.