Inducing orientations on boundary manifolds
Before discussing the boundary of a manifold, perhaps it would be best to look at the case with embedded hypersurfaces in general, and for this I will be following Lee's Intro to Smooth Manifolds.
Let $M$ be a smooth, oriented, $n$-manifold and $S$ be an embedded ($n-1$) hypersurface. Let $\omega$ be an orientation form on $M$, so that $\omega$ is just a nowhere vanishing $n$-form. We would like to find a way of tweaking this $\omega$ to define a nowhere vanishing $n-1$ form on $S$. How do we do this? The key is that since $\omega$ is nowhere vanishing then (without loss of generality) it is always positive; that is, for all $p \in M$, $$ \omega_p\left(\left.\frac{\partial}{\partial x_1}\right|_p,\ldots,\left.\frac{\partial}{\partial x_n}\right|_p\right) >0$$ where the partials form a basis for $T_pM$. If we take $p \in S$, then $T_pS$ will give use $n-1$ vectors to plug into $\omega_p$, so our goal is to find a vector field $v_\text{out}: S \to TM$ such that if $B_p=\left\{\left.\frac{\partial}{\partial x_1}\right|_p, \ldots, \left.\frac{\partial}{\partial x_{n-1}}\right|_p\right\}$ is a basis for $T_pS$ then $\{(v_\text{out})_p\} \cup B_p$ is a basis for $T_pM$. In that case we can define an $n-1$ form $\hat \omega_p$ on $S$ as $$ \hat \omega_p\left(\left.\frac{\partial}{\partial x_1}\right|_p,\ldots,\left.\frac{\partial}{\partial x_{n-1}}\right|_p\right) = \omega_p \left((v_\text{out})_p,\left.\frac{\partial}{\partial x_1}\right|_p,\ldots,\left.\frac{\partial}{\partial x_{n-1}}\right|_p\right)$$ which will also be positive for every $p \in S$ since $\omega$ is positive for every $p \in S \subseteq M$.
So how do we guarantee that $(v_\text{out})_p$ is always "linearly independent" to $B_p$? We simply demand that $(v_\text{out})_p$ never lies in $T_pS$!
Now if we specifically look at boundaries, our discussion above suggests that we find a vector field $N: \partial M \to TM$ such that $N_p \notin T_p\partial M$ (these are usually called inward-pointing vector fields). The Stokes' orientation is to take the $-N$ and define the $\hat \omega_p$ as above.
The other way of defining induced orientations is to do it in terms of the charts, so that it suffices to look at just the upper half plane $$ \mathbb H^n = \{(x_1,\ldots,x_n) \in \mathbb R^n: x_n \geq 0\}.$$ Diffeomorphisms $T: \mathbb H^n \to\mathbb H^n$ with positive Jacobian induce diffeomorphisms $\partial T: \partial \mathbb H^n \to \partial \mathbb H^n$ with positive Jacobian (this is a straightforward but messy exercise with determinants). The orientation form on $\mathbb H^n$ is the standard form $\omega = dx_1 \wedge \cdots \wedge dx_n$, and the vector $-\frac{\partial}{\partial x_n}$ is outward pointing along the boundary. Furthermore, $$\omega\left(-\frac{\partial}{\partial x_n}, \frac{\partial}{\partial x_1}, \ldots, \frac{\partial}{\partial x_{n-1}} \right) = (-1)^n \omega\left(\frac{\partial}{\partial x_1}, \frac{\partial}{\partial x_2}, \ldots, \frac{\partial}{\partial x_{n}} \right)$$ so this suggests that we should take $\hat \omega = (-1)^n dx_1 \wedge \cdots \wedge dx_{n-1}$ to be the induced orientation on $\partial \mathbb H^n$. We then define the induced boundary on $\partial M$ to be $$ [\partial M] = \phi^*[\partial \mathbb H^n]$$ where the square brackets are the equivalence class of orientations and $\phi: M \to \mathbb H^n$ is a coordinate chart of $M$.
In your example, let's use $x$ and $y$ as coordinates for $M$. The outward vector on the boundary should be tangent to M. A suitable choice is $-\partial_x$, considered as a coordinate basis vector on $M$. The minus sign is there because $M$ is on the postive $x$ side, so the outward vector should point to negative $x$ values. The induced orientation on the boundary is therefore $-\partial_x\cdot(dx\wedge dy)= -dy$. So the orientation on the boundary runs from positive $y$ to negative $y$.