My guess would be that a cyclic block is one that is (or maybe is similar to) a companion matrix. This means that for the linear operator $\phi$ defined by $M$ your space has a direct sum decomposition $V=W_1+W_2$ into $\phi$-invariant subspaces, where each $W_i$ is cyclic for $\phi$: there is some vector $w_i$ such that its $\phi$-images $\phi^k(w_i)$ (for $0\leq k<\dim(W_i)$) span the subspace $W_i$.
To answer you question, one could take $v=w_1+w_2$. Then for any polynomial $P$ the projections of $P[\phi](v)$ on $W_1$ and $W_2$ are respectively $P[\phi](w_1)$ and $P[\phi](w_2)$. Let $\mu_1,\mu_2$ be the minimal (or characteristic) polynomials of the cyclic factors. Then $P[\phi](v)=0$ if and only if $P[\phi](w_1)=0$ and $P[\phi](w_2)=0$, which means that both $\mu_1$ and $\mu_2$ divide$~P$; since they are given to be coprime, this means that the product $\mu_1\mu_2$ must divide$~P$. In particular this does not happen for any $P\neq0$ with degree less than $\deg(\mu_1\mu_2)=\deg(\mu_1)+\deg(\mu_2)=\dim(W_1)+\dim(W_2)=\dim(V)$. That means that $\phi^k(w_i)$ for $0\leq k<\dim(V)$ are linearly independent, and $v$ is a cyclic vector for$~V$.
The argument is similar to the one showing that two cyclic subgroups of coprime order in an abelian group have a (direct) sum that is again cyclic, where as generator on can take a sum of the generators of the cyclic factors. And all this is related to the Chinese remainder theorem, which has a version for polynomials as well.