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!