It is known that the tensor product of endofunctors End(C) over a given category C is given by composition and the category of monads Mon(C) over a given category is cartesian. That cartesian product is roughly described in Monad Compositions I (Manes and Mulry) for a given set-indexed family of monads $\mathbf{T_i}=(T_i,\mu_i,\eta_i)$ as $\mathbf{T}=(T,\mu,\eta)$ by the rules:
$\hspace{15em}{TX}=\prod{{T_i}X}$ $\hspace{15em}{\pi_j\eta_X}=\eta_{jX}$ $\hspace{15em}{\pi_j{(X\rightarrow{TY})}}={(TX\rightarrow{T_jX}\rightarrow{T_jY})}$ $\hspace{15em}{\pi_j{(TTX\rightarrow{TX})}}={(TTX\rightarrow{T_jT_jX}\rightarrow{T_jX})}$
where I've omitted the name of the arrows. Is there any reason for this product not to be a tensor product and hence the composition induced by the category End(C) where Mon(C) lives?
Remark: In my version of the paper the third rule was stated as
$\hspace{15em}{\pi_j{(X\rightarrow{TY})}}={(TX\rightarrow{TX_j}\rightarrow{TY_j})}$
which I guess is wrong.
$\mathrm{Mon}(C)$ does not have composition as a tensor. The composite of two monads is not a monad in any natural way unless there is a distributive law between them. Lots of research goes into getting around this fact, e.g. the paper you cite.
For example, for any set $A$ there is a functor $\Delta_A: \mathsf{Set} \to \mathsf{Set}$ which is constant at $A$: it sends all objects to $A$ and all morphisms to $\mathrm{id}_A$. If $A$ has more than one element, then $\Delta_A$ does not admit a monad structure (this is a fun exercise). (It's also true that if $A$ has less than one element, then $\Delta_A$ does not admit a monad structure!)
Now let $T$ be a monad on $\mathsf{Set}$ such that $T(1) \neq 1$. Consider also the monad $\Delta_1$ which is constant at 1. We can compose the underlying functors $T\circ \Delta_1 = \Delta_{T(1)}$ to get the constant functor at the set $T(1)$. But this composite functor does not admit a monad structure.