Let G be a group that can be written as a product of two of its subgroups, that is $$G = MN = \{mn~~|~~m\in M, n\in N\}.$$ Now let $k$ be a field, let $k(N)$ be the dual space of $N$ (here, the dual space means the collection of functions from $N$ to $k$) and $kM$ the group algebra of $M$.
We have that $M$ has an action on $N$ and $N$ also has an action on $M$. These actions induce an action of $k(N)$ on $kM$ and an action of $kM$ on $k(N)$.
I want to know how these actions are possible.
I am not sure what $G$ has to do with this, but let's say $M$ and $N$ are two groups which each have a group action on the other (with no assumption about how $M\curvearrowright N$ respects $N$'s group operation, or vice-versa).
I will use $k^N$ to refer to the algebra of functions $N\to k$ (where $k$ is a field) with pointwise addition and pointwise multiplication of functions as its operations.
Since $M$ acts on $N$, we can say $M$ acts on $k^N$ by algebra automorphisms. Given $f:N\to k$ and any element $m\in M$, the function $m\cdot f$ is defined by $(m\cdot f)(n)=f(m^{-1}n)$ for all $n\in N$. You could call this a sort of "contragredient" action I guess. You can check it respects the pointwise operations on $k^N$.
Given a linear action of a group $M$ on a vector space $V$, that automatically means there is a representation of $kM$ on $V$, by linear extension. That is, $(\sum c_m m)v:=\sum c_m(mv)$. For $V=k^N$ that means
$$ \big((\sum_{m\in M} c_m m)f\big)(n):=\sum_{m\in M} c_m f(m^{-1}n) $$
for all $n\in N$ and $f:N\to k$.