Let me address your last question (and hopefully it will help with clarifying some of your misconceptions):
Are they essentially the same definition except for this weird edge case?
No, not only in that case. Consider e.g. the following set of three vectors in $\mathbb{R}^2$:
$$\mathbf{v}_1=\begin{bmatrix}1\\0\end{bmatrix}, \quad \mathbf{v}_2=\begin{bmatrix}2\\0\end{bmatrix}, \quad \mathbf{v}_3=\begin{bmatrix}0\\1\end{bmatrix}.$$
It's easy to see that this set is linearly dependent according to the standard definition because$$2\mathbf{v}_1+(-1)\mathbf{v}_2+0\mathbf{v}_3=\mathbf{0}.$$However it doesn't satisfy your definition. Although vectors $\mathbf{v}_1$ and $\mathbf{v}_2$ can be expressed as (nontrivial) linear combinations of the other ones, viz. $\mathbf{v}_1=0.5\mathbf{v}_2+0\mathbf{v}_3$ and $\mathbf{v}_2=2\mathbf{v}_1+0\mathbf{v}_3$, we can't do the same with the last vector because the equation$$\mathbf{v}_3=c_1\mathbf{v}_1+c_2\mathbf{v}_2$$clearly has no solutions.
Let me try to describe informally what I think is going on here. The standard definition of linear dependency basically says that there's some dependency somewhere, but not necessarily everywhere, as you seem to believe.
As @AOrtiz already said, one way to think of dependency is that it means redundancy in the given system of vectors. Look at it this way. Given a set of vectors, we may want to construct its span, i.e. the set of all linear combinations of those vectors. If the original set is linearly dependent, then it's redundant in the sense that you can remove some (but not arbitrary!) vectors and still have the same span. The standard definition of linear dependence helps us detect if that's the case.