The composition of the multiplication functors is explicitly given by,

first let,

\\[ F(s)=\underbrace{s \circ s \circ \cdots \circ s}\_{m \text{ times }}= m \\\\
G(s)=\underbrace{s \circ s \circ \cdots \circ s}\_{n \text{ times }}= n \\\\
s \circ s \circ \cdots \circ s = x
\\]

then,

\\[
(G \circ F)(s \circ s \circ \cdots \circ s) \\\\
= G(F(s \circ s \circ \cdots \circ s)) \\\\
= G(F(s) \circ F(s) \circ \cdots \circ F(s)) \\\\
= G(m \circ m \circ \cdots m) \\]

\\[
= G(\underbrace{s \circ s \circ \cdots \circ s}\_{m }) \circ G(\underbrace{s \circ s \circ \cdots \circ s}\_{m }) \circ \cdots G(\underbrace{s \circ s \circ \cdots \circ s}\_{m }) \\\\
= \underbrace{G(s) \circ G(s) \circ \cdots \circ G(s)}\_{m } \circ \underbrace{G(s) \circ G(s) \circ \cdots \circ G(s)}\_{m } \circ \cdots \underbrace{G(s) \circ G(s) \circ \cdots \circ G(s)}\_{m }
\\]

\\[
= \underbrace{n \circ n \circ \cdots \circ n}\_{m} \circ
\underbrace{n \circ n \circ \cdots \circ n}\_{m } \circ \cdots
\underbrace{n \circ n \circ \cdots \circ n}\_{m } \\]
\\[
= n \ast m \ast x,
\\]

which follows simply from the functor laws!