You're right that if the surface has to fit into 3D, and if it is non-self-intersecting, then it has to be orientable.
But if you allow the 2D surface to self-intersect or go to the 4th dimension, there are many counterexamples. Klein Bottle is the simplest one.
Take a bottle, heat it up, and push the throat through the body, to connect it with a hole that you prepared at the bottom of the bottle. This manifold is equivalent to a $Z_2$ orbifold of a torus - where the $Z_2$ generator shifts by half a period in one direction of the torus and reflects the other direction of the torus.
Your rule (1) fails because there is no well-defined interior and exterior. Indeed, if you make trips around the Klein bottle that reverse the orientation, it inevitably exchanges the interior with the exterior as well. Note that if you enter the hole at the bottom of the bottle (see the picture), you're still outside, but as you travel through the throat, it becomes clear that you have gotten inside the "object" as well, so there's no distinction between the exterior and interior.
The Klein bottle is self-intersecting if embedded into three dimensions. Alternatively, you may avoid the intersections if one of the pieces of the surface that would intersect each other are shifted in a new, fourth dimension of space.
To answer the question directly, yes, in terms of forms,
\begin{align}
\int_S\mathbf{F}\cdot d\mathbf{a}&=\int_S\iota_S^*\left(F_1\,dy\wedge dz+F_2\,dz\wedge dx+F_3\,dx\wedge dy\right)
\end{align}
where $\iota_S:S\to\Bbb{R}^3$ simply is the inclusion, and we're pulling back a $2$-form on $\Bbb{R}^3$ to a $2$-form on $S$.
It may be more instructive to look at the situation in generality. So, let $(M,g)$ be an $n$-dimensional Riemannian manifold which is oriented with volume form $\mu_g$, which we may on occasion write as $dV_g$ (it is also common to write this as $\mu_M$ when the metric $g$ is understood), even though it is not the exterior derivative of anything. Recall that $\mu_g$ is characterized by the fact that for any $p\in M$ and any positively-oriented ordered, orthonormal basis $\{\xi_1,\dots, \xi_n\}$ of $T_pM$, we have
\begin{align}
\mu_g(\xi_1,\dots, \xi_n)&=1.
\end{align}
Let $S\subset M$ be a smooth $(n-1)$-dimensional embedded submanifold (i.e a smooth hypersurface), and equip it with the induced Riemannian metric $\iota^*g$. Suppose also that there is a smooth unit normal $N$ defined on $S$ (i.e $N:S\to TM$ is smooth and for each $p\in S$, $N(p)\in (T_pS)^{\perp}$). With this, one typically defines an orientation on $S$ by saying that an ordered basis $\{\xi_1,\dots, \xi_{n-1}\}$ of $T_pS$ is positively-oriented if and only if $\{N(p), \xi_1,\dots, \xi_{n-1}\}$ is a positively-oriented ordered basis for $T_pM$ (i.e $\mu_g$ evaluated on the $n$ vectors in this order is strictly positive).
Since $(S,\iota^*g)$ is an oriented Riemannian manifold in its own right, it has its own volume form, $\mu_{\iota^*g}$ (other common notations include $\mu_S$ or $dS$ or $dA$ or $da$ or $d^{n-1}V$ or $dV_{\iota^*g}$ etc). Now, we can investigate the relationship between the induced volume form on $S$ and the volume form on $M$. For this, let $F:S\to TM$ be a smooth map such that for each $p\in S$, $F(p)\in T_pM$ (i.e a field of vectors defined on $S$, but tangent to $M$; we're not requiring it to be tangent to $S$). Let us also define the normal and tangential components of $F$ in the usual way:
\begin{align}
F_{\perp}&:= g(F,N)\,N\equiv \langle F,N\rangle N\quad \text{and}\quad F_{\parallel}:= F-F_{\perp}
\end{align}
Let $p\in M$ and let $\{\xi_1,\dots, \xi_{n-1}\}$ be any basis for $T_pM$. We now look at the interior product:
\begin{align}
(F_{\parallel}\,\lrcorner \,\mu_g) (\xi_1,\dots, \xi_{n-1})&:= \mu_g(F_{\parallel}(p), \xi_1,\dots, \xi_{n-1})=0,
\end{align}
where the last equality is because $F_{\parallel}(p)\in T_pS$ is tangent to $S$, meaning it can be written as a linear combination of the $\xi_i$'s and hence each summand vanishes (because the same $\xi$ appears twice in the evaluation of $\mu_g$). Therefore, the interior product of $F_{\parallel}$ with $\mu_g$ vanishes, or rearrranging this equation, we have
\begin{align}
F\lrcorner\,\, \mu_g &= \langle F,N\rangle\, (N\lrcorner\,\, \mu_g) =\langle F,N\rangle\,\, \mu_{\iota^*g},\tag{$*$}
\end{align}
where the last equality can be easily verified by the characterization of Riemannian volume forms I mentioned above: take any $p\in S$ and any positively-oriented, ordered, orthonormal basis $\{\xi_1,\dots, \xi_{n-1}\}$ of $T_pS$. Then,
\begin{align}
(N\,\lrcorner\,\,\mu_{g})(\xi_1,\dots, \xi_{n-1})&= \mu_g(N(p),\xi_1,\dots, \xi_{n-1})=+1,
\end{align}
where the final equality is because $\{N(p),\xi_1,\dots, \xi_{n-1}\}$ is an ordered orthonormal basis for $T_pM$, which is also positively-oriented by definition of the orientation on $S$. Note also, that $\mu_{\iota^*g}$ is the unique such form. Hence, $\mu_{\iota^*g}= N\,\lrcorner \, \mu_g$, as claimed above in $(*)$.
Now, let us connect this all back to your question. We have by $(*)$ that
\begin{align}
\mathbf{F}\cdot d\mathbf{a}&\equiv\langle F,N\rangle\, \mu_{\iota^*g}=F\,\lrcorner\,\mu_g\equiv F\,\lrcorner \,\, dV_g. \tag{$**$}
\end{align}
Here, the first and last equal signs are just notation, only the middle one is significant. Hence,
\begin{align}
\int_S\,\mathbf{F}\cdot d\mathbf{a}&= \int_S F\,\lrcorner\,dV_g.
\end{align}
If we want, we can now extract the formula in terms of local coordinates. Let $(U,x= (x^1,\dots, x^n))$ be a positively-oriented coordinate chart. Then, we can write
\begin{align}
\mu_g\equiv dV_g&= \sqrt{|\det (g_{ij})|}\,dx^1\wedge \cdots \wedge dx^n\quad \text{and}\quad F=\sum_{i=1}^nF^i\frac{\partial}{\partial x^i}
\end{align}
Here, $F^i:S\cap U\to\Bbb{R}$ are smooth functions. So, the interior product is given by
\begin{align}
F\,\lrcorner\,dV_g&=\left(\sum_{i=1}^nF^i\frac{\partial}{\partial x^i}\right)\,\lrcorner \sqrt{|\det g|}\,dx^1\wedge \cdots \wedge dx^n\\
&=\sum_{i=1}^n \sqrt{|\det g|}F^i \left(\frac{\partial}{\partial x^i}\,\lrcorner\,dx^1\wedge \cdots\wedge dx^n\right)\\
&=\sum_{i=1}^n\sqrt{|\det g|}F^i\, (-1)^{i-1}\,dx^1\wedge \cdots \wedge \widehat{dx^i}\wedge \cdots\wedge dx^n
\end{align}
(ok here I've been a little lazy with the notation regarding restrictions/pullbacks to $S\cap U$). Note btw that in the very special case of $\Bbb{R}^3$ with its standard Riemannian metric, and in Cartesian coordinates $(x^1,x^2,x^3)=(x,y,z)$, we have $\sqrt{|\det g_{ij}|}=1$, and the summation fully expanded out is
\begin{align}
F^1\,dx^2\wedge dx^3-F^2\,dx^1\wedge dx^3+F_3\,dx^1\wedge dx^2&\equiv F^1\,dy\wedge dz+F^2\,dz\wedge dx+F^3\,dx\wedge dy,
\end{align}
which is exactly what you found.
Best Answer
What you are looking for is Alexander's duality theorem. One of its immediate corollaries is that compact non-orientble surfaces do not embed in three-space.