Let $L$ be a finite Galois group that's solvable. So by definition there exists a chain of normal subgroups of $L$ such that $1 = G_{0} \trianglelefteq G_{1} \trianglelefteq \dots \trianglelefteq G_{n}=L $ where all the $G_{i+1}/G_{i}$ are abelian.
Now, it's said that one can assume the $G_{i+1}/G_{i}$ to be cyclic by the structure theorem for finite abelian groups.
Why exactly is that?
Take a finite group $G$ and a normal subgroup $H_0\subseteq G$. Then there is a one-to-one correspondence between normal subgroups of $G/H_0$ and normal subgroups of $G$ that contain $H_0$.
If $G/H_0$ is abelian, then by the structure theorem of finite abelian groups, we have that $G/H_0$ is isomorphic to a finite product of finite, cyclic groups. Specifically, this means that unless $G = H_0$, $G/H_0$ has non-trivial cyclic subgroups. Pick one such subgroup and let $H_1$ be the corresponding normal subgroup of $G$. Then we have shown from $H_0 \trianglelefteq G$ and $G/H_0$ abelian that there is a $H_1\subseteq G$ such that $H_0 \trianglelefteq H_1 \trianglelefteq G$, and $H_1/H_0$ is cyclic. By induction on, for instance, the index of $H_i$ in $G$, we get a finite chain of subgroups $H_0 \trianglelefteq H_1 \trianglelefteq H_2\trianglelefteq \cdots \trianglelefteq H_n = G$ such that each $H_{i+1}/H_i$ is cyclic
A quote from wikipedia: A group $G$ is called solvable if it has a subnormal series whose factor groups are all abelian. For finite groups, an equivalent definition is that a solvable group is a group with a composition series all of whose factors are cyclic groups of prime order. This is equivalent because a finite group has finite composition length, and every simple abelian group is cyclic of prime order.
For more details see the proof here.