How is the normal field for a 2-dimensional submanifold of $\mathbb R^3$ defined? Let $M$ be a 2-dimensional embedded $C^1$-submanifold of $\mathbb R^3$, i.e. a subset of $\mathbb R^3$ such that for all $x\in M$ there is a 2-dimensional chart$^1$ $(U,\phi)$ of $M$ with $x\in\phi(U)$. In particular, for all $u\in U$ and $x:=\phi(u)$, $N_x\phi(U)$ is 1-dimensional and hence there is an unique $\nu_{\phi(U)}(x)\in N_x\phi(U)$ of unit length with $$\det\left({\rm D}\phi(u),\nu_{\phi(U)}(\phi(u))\right)>0\tag2.$$ To be precise, $$\nu_{\phi(U)}(x)=\frac{\partial_1\phi(u)\times\partial_2\phi(u)}{\left|\partial_1\phi(u)\times\partial_2\phi(u)\right|}.\tag3$$

Question 1: In general there is a countable system $\mathfrak A$ of 2-dimensional charts with $M\subseteq\bigcup_{(U,\:\phi)\in\mathfrak A}\phi(U)$. Are we able to impose a condition on $M$ ensuring that there exists a choice of $\mathfrak A$ such that each chart in $\mathfrak A$ satisfies $(2)$? I guess this is related to the notion of "orientability", but how is it defined here? If this problem is solved, can we immediately conclude the existence of $\nu_M$ by defining $\nu_M(x)=\nu_{\phi(U)}(x)$ for $x\in M$ and $(U,\phi)\in\mathfrak A$ with $x\in\phi(U)$? Is this well-defined, unique and continuous?
Question 2: How can we generalize the existence of $\nu_M$ to more general $M\subseteq\mathbb R^3$? I don't want to get too general, but I need to include the case of a cube formed from six squares or a box sitting on a table. So, maybe something like $M$ being the finite union of "piecewise $C^1$" embedded $C^1$-submanifolds $M_i$ "possibly with boundary". However, I'd need some help how exactly "piecewise $C^1$ and "with boundary" need to be defined here. And I wonder whether we need to assume that each $M_i$ is a closed set.


$^1$ i.e. $U\subseteq\mathbb R^2$ is open, $\phi:U\to\mathbb R^3$ is an immersion and a topological embedding of $U$ into $M$ and $\Phi(U)$ is an open subset of $M$ (equipped with the subspace topology).
$^2$ $v\in\mathbb R^3$ is called tangent vector on $M$ at $x\in M$ if there is a $\varepsilon>0$ and a $\gamma\in C^1((-\varepsilon,\varepsilon),M)$ with $\gamma(0)=x$ and $\gamma'(0)=v$.
 A: Answer to the original question: The answer to Question 1 is "No". The Möbius band is a counterexample; indeed, if Question 1 held then $M$ would be orientable, and any nonorientable surface in $\mathbb R^3$ is a counterexample. But it's worse than that, even an orientable surface can have a system of charts which fails to satisfy Question 1. All you need is to have one chart $(U,\phi_1)$ for which $U \subset \mathbb R^2$ is the open unit disc $U=\{(x,y) \mid x^2+y^2<1\}$, and then you can expand your system by adding one more chart $(U,\phi_2)$ such that $\phi_2(x,y) = \phi_1(-x,y)$; the normal vectors corresponding to these two charts will point in opposite directions.
And because the answer to Question 1 is "No", Question 2 makes no sense.
Added to address edited remarks concerning orientability: What you're missing, perhaps, is a definition of an orientable system of charts. The requirement for the system of charts to be orientable is that if two charts $(U_1,\phi_1)$, $(U_2,\phi_2)$ have a common overlap $p \in M$, with $p = \phi_1(q_1) = \phi_2(q_2)$, then the Jacobian determinant of the derivative of the overlap map $D_{q_1} \phi_2^{-1} \circ \phi_1$ must be positive. The existence of such a system of charts is equivalent to the existence of a system of well-defined, unique, and continuous system of normal vectors $\nu$.
Let me add that the concept of "outward" is not well-defined in general: everything in your two questions is solely about orientability. Where the concept of "outward" becomes relevant is when $M$ is compact and connected, in which case one applies other tools to construct the "outward" normal. But, I think that's a topic for a different question.
