The Product Neighborhood Theorem states that if $N\subseteq M$ is a smooth compact submanifold without boundary of codimension $n$ and there is a trivialization of the normal bundle (wrt. some smooth metric) of $N$ in $M$, then some neighborhood of $N$ is diffeomorphic to $N\times \mathbb{R}^n$ s.t. the diffeomorphism takes the framing vectors to the canonical basis of $\mathbb{R}^n$.
Is the same true if $N,M$ are smooth manifolds with boundary and $N$ is a neat submanifold of $M$?
For the proof, in most texts I found the idea to map $(n,x)\in N\times\mathbb{R}^n$ to $\varphi(1)$ for a geodesics $\varphi$ starting in $n$ with tangent vector $\epsilon x$ for $\epsilon$ small enough. This probably doesn't work on the boundary: for example, if $M$ is the closed unit 2-disc and $N:=\{0\}\times [-1,1]$ with the horizontal framing $(1,0)$ then I don't see how to generalize the above construction. I can hardly have a geodesics in $M$ starting in $(0,1)$ in the $(1,0)$ direction.
Remark: What confuses me is that the boundary-version of PNT is used in the proof of the Thom-Pontryiagin construction -- namely, that $[M,S^n]$ is isomorphic to the framed cobordism group $\Omega^{fr}_{n;M}$ (a framed cobordism has boundary) -- but in all books, I have only seen the statement of PNT for the boundaryless case.