How can we show that this normal field is "outward pointing"? Let $d\in\mathbb N$, $\alpha\in\mathbb N$ and $M$ be a $d$-dimensional embedded $C^\alpha$-submanifold of $\mathbb R^d$ with boundary.

How can we show that there is a unique $\nu_M:\partial M\to\mathbb R^d$ with$^1$ $$\nu_M(x)\in N_x\:\partial M,\tag1$$ $\left\|\nu_M(x)\right\|=1$ and $$\exists\varepsilon>0:\forall t\in(0,\varepsilon):x+t\nu_M(x)\not\in M\tag2$$ for all $x\in\partial M$?

Let $\phi$ be a $C^1$-diffeomorphism$^1$ from an open subset $\Omega$ of $M$ onto an open subset of $\mathbb H^d:=\mathbb R^{d-1}\times[0,\infty)$. By definition of $C^1$-differentiability, $$\phi_d=\left.\psi\right|_\Omega\tag3$$ for some $\psi\in C^1(O)$ for some open subset $O$ of $\mathbb R^d$ with $\Omega\subseteq O$. Since $\phi(\Omega)\subseteq\mathbb H^d$, $$\psi(\Omega)\subseteq[0,\infty)\tag4.$$ Moreover, since $\phi(\Omega\cap\partial M)=\phi(\Omega)\cap\partial\mathbb H^d\subseteq\partial\mathbb H^d$, $$\psi(\Omega\cap\partial M)\subseteq\{0\}\tag5.$$ Assume $\Omega\cap\partial M\ne\emptyset$ and hence $$\psi(\Omega\cap\partial M)=\{0\}\tag6.$$ Let $x\in\Omega\cap\partial M$. Note that, if $h\in\mathbb R^d$ with $$\langle{\rm D}\phi(x)h,e_d\rangle>0\tag7,$$ then (since $\psi(x)=0$) $$\frac{\psi(x+th)}t\xrightarrow{t\to0}{\rm D}\psi(x)h=\langle{\rm D}\phi(x)h,e_d\rangle>0\tag8$$ by the chain rule and hence $$\psi(x+th)<0\;\;\;\text{for all }t\in(-\varepsilon_1,0)\tag9$$ for some $\varepsilon_1>0$ and $$\psi(x+th)>0\;\;\;\text{for all }t\in(0,\varepsilon_2)\tag{10}$$ for some $\varepsilon_2>0$.
Now let $$\mu_M(x):={{\rm D}\phi(x)}^\ast e_d=\nabla\phi_d(x)=\nabla\psi(x)\tag{11},$$ where $(e_1,\ldots,e_d)$ denotes the standard basis of $\mathbb R^d$. Then $$N_x\:\partial M=\mathbb R\mu_M(x)\tag{12}.$$ Let $$\nu_M(x):=\frac{\mu_M(x)}{\left\|\mu_M(x)\right\|}.$$ Then, obviously, $$\langle{\rm D}\phi(x)\mu_M(x),e_d\rangle=\left\|{{\rm D}\phi(x)}^\ast e_d\right\|^2>0\tag{13}$$ and hence $\nu_M(x)$ satisfies $(2)$.

How do we need to argue that $\nu_M(x)$ is actually uniquely determined by $(2)$? And how can we show that there is a tubular neighborhood $U$ of $\partial M$ so that $\nu_M$ can be extended to a function in $C^{\alpha-1}(U,\mathbb R^d)$?

I think we should first try to extend $\mu_M$. In order to do that, let $((\Omega_i,\phi_i))_{i\in I}$ be a $C^\alpha$-atlas of $M$ for some nonempty set $I$. As before, $$\langle\phi_i,e_d\rangle=\left.\psi_i\right|_\Omega\tag{14}$$ for some $\psi_i\in C^\alpha(O_i)$ for some open subset $O_i$ of $\mathbb R^d$ with $\Omega_i\subseteq O_i$ for all $i\in I$. Since $\Omega_i$ is an open subset of $M$, we can assume that $$\Omega_i=O_i\cap M\;\;\;\text{for all }i\in Itag{15}.$$ Let $$\mu_i(x):={{\rm D}\phi_i(x)}^\ast e_d=\nabla\psi_i(x)\tag{16}$$ and $$\nu_i(x):=\frac{\mu_i(x)}{\left\|\mu_i(x)\right\|}$$ for $x\in O_i$ and $i\in I$. Let $O:=\bigcup_{i\in I}$.

Now, the idea is to take $\rho_i\in C_c^\infty(O)$ with $\left.\rho_i\right|_{\partial M}\ge0$ and $\operatorname{supp}\rho_i\subseteq O_i$ for $i\in I$ with $$\partial M\subseteq\left\{\sum_{i\in I}\rho_i=1\right\}\tag{17}.$$ Which assumption do we need to find $(\rho_i)_{i\in I}$? And how can we conclude that $$\overline\mu_M(x):=\sum_{i\in I}\rho_i(x)\mu_M(x)\;\;\;\text{for }x\in O$$ is actually a well-defined extension of $\mu_M$?

I'm unsure how we need to argue that, if $x\in\Omega_i\cap\Omega_j$ for some $i,j\in I$ with $i\ne j$, $\mu_i(x)=\mu_j(x)$. And don't we "count them twice" in $\overline\mu_M(x)$?

$T_x\:\partial M$ and $N_x\:\partial M$ denote the tangent and normal space of $\partial M$ at $x$, respectively.
 A: For each point point $x\in\partial M$, $T_x\partial M$ is a $d-1$ dimensional subspace and thus $N_x\partial M$ contains exactly exactly two vectors of unit norm. Thus, unit normal vector fields are unique iff $(2)$ holds for exactly one of them. You already have everything needed to prove this, since in local coordinates exactly one of them satisfies the inequality $(7)$.
It won't always be possible to extend a normal vector field $\nu$, depending on what definition of submanifold with boundary you are using. For instance, the set
$$
S=\{(x,y)\in\mathbb{R}^2:x\ge 0,y\ge 0,(x,y)\neq(0,0)\}
$$
Is an embedded submanifold with boundary, but there is no extension of the normal vector field near the origin.
We can, however, always extend $\nu$ provided $\partial M$ is closed $\mathbb{R}^d$. In this case, we can cover $\partial M$ with a collection of boundary charts $(\phi_i,\Omega_i)$ of $\partial M$ which, together with $\mathbb{R^d}\setminus\partial M$, form an open cover of  $\mathbb{R}^d$. We can then define a partition of unity $\rho_i$ subbordinate to this cover. If we can define a local extension $\nu_i:\Omega_i\to\mathbb{R}^n$ for each $i$,  defined in coordinates by $\nu_i(x^1,\dots,x^d)=\nu(x^1\dots,x^{d-1},0)$, then we may define a global extension $\widetilde{\nu}(x)=\sum_i\rho_i\nu_i(x)$.
