From page 15 of Lambda Calculus and Combinators an Introduction:
Note 1.34 If $M \equiv aM_1 \ldots M_n$ where $a$ is an atom, and $M \triangleright_\beta N$, then $N$ must have form
$$ N \equiv a N_1 \ldots N_n $$
where $M_i \triangleright_\beta N_i$ for $i = 1, \ldots, n$. To see this, note that $M$ is really
$$ (( \ldots ((aM_1)M_2) \ldots)M_n) $$
when its parentheses are fully shown; hence each $\beta$-redex in $M$ must be in $M_i$. Also the same holds for each subterm $\lambda x. P$ whose bound variable might be changed in the reduction of$M$.
Question: Why is the following not a counter-example?
First let $M_1 = m_1 m_2 m_3$ for atomic constants $m_1$, $m_2$, $m_3$. Then let $N_1 = m_1$ and $N_2 = m_2 m_3$. Then:
$$ aM_1 \equiv a (m_1 m_2 m_3) \equiv (((am_1)m_2)m_3) \equiv ((a m_1)m_2 m_3) \triangleright_\beta a N_1 N_2 $$
Here, if I am correct, then we have a case where
$$ a M_1 \triangleright_\beta a N_1 N_2 $$
which seems to contradict the note.