Let $M$ be a smooth manifold with boundary, endowed with a smooth Riemannian metric $g$.
Suppose $g$ is flat, and let $p \in \partial M$.
Is there an open neighbourhood of $p$ which is isometric to an open subset of the standard half-space $\mathbb{H}^n=\{ (x_1,...,x_n)| x_n \ge 0\}$, endowed with the standard Euclidean metric?
One approach would to be to use the exponential map at $p$ (which is a local isometry in the case with no boundary). However, I am not sure where $\exp_p$ is defined in this case.
I guess its domain will be the subset of inward-pointing vectors in $T_pM$.
Edit:
As seen by the example $M = \{ x \in \mathbb R^2 \,:\, |x| \leq 1 \} $, the suggetsion to use the exponential map does not work. The problem is that to cover the boundary, we need to conside geodesics with initial velocities in $T_p\partial M$. But, if we consider these geodesics in the ambient space $\mathbb{R}^2$, then they are standard straight lines (which are tangent to the unit circle), hence do not lie in $M$ for any positive time.
So, indeed, the boundary "cannot be straightened out".
I would say no, because you cannot assure that the boundary will be straight. Consider the disk $$ M = \{ x \in \mathbb R^2 \,:\, |x| \leq 1 \}\,. $$ It is clearly flat, but because the only isometries in $\mathbb R^2$ are translations and rotations you cannot straighten out the boundary, not even locally.
I would suggest the following amendment: every flat Riemannian manifold with boundary is locally isometric to an open subset together with parts of its smooth boundary.