Let me recall the definition of a rational $G$-module from M. Brions notes Introduction to actions of algebraic groups (Def. 1.6)
Let $G$ be an affine group scheme over $\mathbb{C}$. A rational $G$-module is an action of $G$ on a (not necesseraly finite dimensional) $\mathbb{C}$-vector space V such that every $v \in V$ is contained in a finite dimensional $G$-stable subspace of $V$ on which $G$ acts algebraically.
Just after this definition Brion states that if $G$ is irreducible and not trivial, then the action of $G$ on the rational function field $\mathbb{C}(G)$ is an example of a $G$-module that is not rational.
The first point I do not understand here is the following: Brion says that $G$ operates on $\mathbb{C}(G)$ via left multiplication. On the coordinate ring $\mathbb{C}[G]$ the group $G$ operates via right multiplication: $(g.f(x)) = f(xg)$ for $g,x \in G$ and $f \in \mathbb{C}[G]$. So I would guess that one would try to extend this action to $\mathbb{C}[G]$. However it is not clear to me why this works in general. So the first question is: How does $G$ act on $C(G)$?
Second, I would like to see a concrete example where this action is not rational, ideally in an easy case such as $G = \mathbb{G}_a$.
As I understand it, it seems that the existence of non-rational $G$-modules would contradict the fundamental lemma on representations (every representation is the finite unions of its finite-dimensional subrepresentations), see e.g. Corollary 4.7 in Milnes book Algebraic Groups. Probably I am misunderstanding something fundamental here. I appreciate clarifications.
Let us for the moment forget all about algebraic geometry and just consider a group $G$ and a $G$-module $V$. The definition of being rational makes sense in this context, so let us just find an example of a non-rational $G$-module which will work whenever $G$ is infinite (so it will also work for whatever algebraic group you choose as long as it is infinite).
It turns out that we can just take the regular $G$-module (note that this term tends to mean something slightly different when we consider $G$ as an algebraic group though). So let $V$ be the vector space with basis $\{v_g\mid g\in G\}$ and define the action of $G$ by letting $gv_x = v_{gx}$.
I claim that this is not a rational $G$-module. To see this note that the submodule generated by any $v_g$ is all of $V$, so none of these can be contained in a finite dimensional $G$-stable subspace (and in fact, this module has no non-trivial finite dimensional submodules).