Actually, let's even forget about volume forms.
Start with the manifold ${\bf R}^{N}$, regard it as a vector space (in the future, as a tangent space at a point of a manifold), and choose a basis:
$$
{\bf E} = \{{\bf e}^{k} \; : \; k = 1, 2, \ldots, N\}
$$
(in the future, a frame on the manifold), with the qualification that the basis is ordered (say, by the indices $k$). (For $N$, the order ${\bf e}^{1} = (1, 0), {\bf e}^{2} = (0, 1)$ corresponds to what we know colloquially as the counterclockwise orientation; to help the visual, traverse the points points: ${\bf 0} = (0,0), {\bf e}^{1} = (1, 0), {\bf e}^{2} = (0, 1)$, in that order).
A linear transformation $A$ which transforms one basis (frame) to another will, therefore, be invertible, hence has a nonzero determinant. The transformation with a positive determinant is said to $\textit{preserve orientation}$ with a negative determinant, to $\textit{change orientation}$. (Test this on the 2-D example I gave above.)
Two frames have the same orientation if one is transformed into the other by a transformation with a positive determinant. This establishes an equivalence relation on the set of frames, with two equivalence classes. These classes are called orientations.
Hope this sheds light on it.
A source: Mathematical Analysis II, V.A. Zorich
The two examples shown do explicitly use the definition. But the argument is not an "equations" argument, so perhaps you won't be satisfied. I'll add some details all the same.
Example 1 is of an atlas consisting of just one neighborhood. The requirement in the definition regarding "two neighborhoods of this family" holds vacuously, because there do not exist two neighborhoods in that atlas. No equations are therefore necessary. As the author writes, the surface is "trivially orientable", because the definition holds vacuously.
Example 2 is of an atlas consisting of exactly two neighborhoods, one with parameters $(u,v)$ and the other with parameters $(\bar u,\bar v)$. The requirement in the definition "two neighborhoods of this family" therefore needs to be checked for these two neighborhoods. Now, before he checks that, there is a preliminary task to be carried out, which he describes in the sentence starting "If the Jacobian of the coordinate change at $p$ is negative... interchange $u$ and $v$...". In other words, if
$$\text{det}\pmatrix{\partial u / \partial \bar u & \partial u / \partial \bar v \\ \partial v / \partial \bar u & \partial v / \partial \bar v} < 0
$$
then, swapping rows 1 and 2 with each other, we obtain
$$\text{det}\pmatrix{\partial v / \partial \bar u & \partial v / \partial \bar v \\ \partial w / \partial \bar u & \partial w / \partial \bar v} > 0
$$
Thus, we now have an atlas with two coordinate charts having connected intersection $W$, and having positive Jacobian at one point $p \in W$. But the definition requires that the Jacobian be positive at all points of $W$. This follows from a topological fact: the Jacobian determinant is a nonzero continuous function on the connected space $W$ and therefore it has constant sign on $W$. Since that sign is positive at one point, namely $p$, it is positive at every point.
Best Answer
Let $\varphi_i:V_i\to\mathbb R^n$ be the given coordinate maps. Then the map $f:V_1\cap V_2\to\mathbb R$, given by $$f(x)= \det(d(\varphi_2^{-1}\varphi_1)_x)$$ is continuous, and $f(x)\neq0$ for all $x\in V_1\cap V_2$ (since $d(\varphi_2^{-1}\varphi_1)_x$ is always invertible). Since $V_1\cap V_2$ is assumed to be connected, the image of $f$ is connected, hence must lie in some connected subset of $\mathbb R\setminus\{0\}$. Thus the image of $f$ is a subset of either $(-\infty,0)$ or $(0,\infty)$, i.e., there is no sign change.