Let $\mathbb R^m_{-}$ be the closed left half space and let $f: U \to V$ be a diffeomorphism between relatively open sets in $\mathbb R^m_{-}$, that is, there exists an extension $\hat f: \hat U \to \hat V$ which is a smooth map between open sets $U, V \in \mathbb R^m$ such that $U = \hat U \cap \mathbb R^m_{-}$ and $V = \hat V \cap \mathbb R^m_{-}$ and $\hat f \equiv f$ on $U$. And further $f$ is bijective and has smooth inverse (in sense above).
1) We define the differential of $f$ at a point $x \in U$ to be equal the differential of $\hat f$ at $x$. Why is the differential of $f$ an isomorphism for all points in $U$? For usual diffeomorphism between open sets of Euclidean space this follows from the chain rule. How to argue precisely here?
2) A boundary point of $U$ is a point $(0,y) \in U$ with $y \in \mathbb R^m$. How does the inverse function theorem show that boundary points get mapped to boundary points?
1) Since $\hat{f}$ is a diffeomorphism, it has an inverse $\hat{f}^{-1}$. Computing the chain rule for $\hat{f} \circ \hat{f}^{-1} = \textrm{id}$ and $\hat{f}^{-1} \circ f = \textrm{id}$ shows that $(D\hat{f})^{-1} = D\hat{f}^{-1}$ i.e $Df(q): T_q\mathbb{R}^m \to T_{f(q)} \mathbb{R}^m$ is an invertible linear map on finite dimensional vector spaces and so it is an isomorphism.
2) Let $\hat{f}(0,y) = p$. I don't think you don't need the Inverse function theorem if $\hat{f}$ is already a diffeomorphism. In any case we have $D\hat{f}(0,y)$ is an isomorphism and so if $p \not \in \partial \hat{V}$ then we would be saying $ T_{(0,y)} \partial U \cong \mathbb{R}^{m-1}$ is vector space isomorphic to $T_pV \cong \mathbb{R}^m$ which is not true.