I would prefer you state your definition of linear independence thusly:
Definition: The subset $\{v_1,\dots,v_n\}\subset V$ is linearly independent if whenever $a_1,\dots,a_n\in F$ and$$a_1v_1+\dots+a_nv_n = 0,$$then $a_1 = \dots = a_n = 0$.
Let's see how your intuition breaks down:
Definition: A set $A=\{v_1,\dots,v_n\}\subset V$ is linearly dependent if for each $v_i\in A$ there is a nontrivial linear combination$$a_1v_1+\dots+\widehat{a_iv_i}+\dots+a_nv_n = v_i,$$where the notation $a_1v_1+\dots+\widehat{a_iv_i}+\dots+a_nv_n$ means that $a_iv_i$ is excluded from the sum.
This says that every vector in the set $A$ can be expressed as a nontrivial linear combination of the other vectors. Well, what if we consider the set $A=\{e_1, 0\}$, where $e_1 = \begin{bmatrix} 1 \\ 0\end{bmatrix}$ is a column vector in $\Bbb R^2$, and $0$ is the zero vector. Then, according to our definition, this set $A$ is not dependent, since we can't express $0$ as a nontrivial linear combination of $e_1$. However, we expect this to be dependent because of course $0$ does depend on $e_1$, as in $0 = 0e_1$. [As I said in my comment to the OP below, I don't like how I originally phrased this—I would rather explain the intuition for why $\{e_1,0\}$ is dependent solely in terms of the redundancy that $0$ brings to this set.]
A better intuition for linear independence is that a set is linearly independent if we are specifying a minimal amount of information for the space it spans. That is, we can always consider the span of a set of vectors $A\subset V$. If we specify the minimal amount of information to achieve the span of $A$, then there are no redundancies: the vectors are independent. So dependent sets should be ones where we can find redundant information leftover.
To be concrete about the idea of how dependence $\leftrightarrow$ redundancies, consider the set $\{e_1,0\}$ again; this time consider its span too, i.e., $\{a_1e_1 + a_20:a_1,a_2\in F\}$. The $0$ vector is redundant because $\operatorname{span}(\{e_1,0\}) = \operatorname{span}(\{e_1\})$. Thus the $0$ vector is redundant, and the set is dependent.
On the other hand, if a set is independent, like $A=\{e_1,e_2\}\subset\Bbb R^2$, then we should not be able to remove even one vector from the set $A$ without changing $\operatorname{span}(A)$. This bears itself out here of course—reinforcing the intuition that independence $\leftrightarrow$ specifying the minimal amount of information.