The theorem states that if $(v_1, ...,v_m)$ is linearly dependent in $V$ and $v_1 \neq 0$ then there exists $j \in \{2,...,m\}$ such that $v_j \in span(v_1,...,v_{j-1})$.
If $(v_1, v_2, ..., v_m)$ are linearly dependent, then by definition of linear dependence, at least one of these vectors can be expressed as a linear combination of remaining vectors.
To be more precise, vectors are linearly dependent if not all scalars $a_i$ have to be zero in this equality:
$$a_1 v_1 + a_2 v_2 + ... + a_n v_n = 0$$
For instance if $0v_1+0v_2+3v_3=0$, then $v_2=-3v_3$. So you just pick the vector $v_j$ that's associated with non-zero $a_j$, subtract everything else from both sides and get $v_j$ expressed as a linear combination of remaining vectors. Therefore $v_j \in span(v_1,...,v_{j-1})$, by definition of span.
I just don't understand the requirement that $v_1 \ne 0$. It always works, no matter if $v_1$ is zero or not.