classical mechanics - What is the momentum mapping for a left translation group action?

This question is a follow-up to a previous question. In this survey of Symplectic Geometry by Arnol'd and Givental, the example of a constant-speed moving particle considered in the previous question was briefly mentioned on Page 62 as follows:

The action of the group by left translations on its cotangent bundle is Poisson. The corresponding momentum mapping $P:T^*G \to \mathfrak{g}^*$ coincides with the right translation of covectors to the identity element of the group.

We can take $G = \mathbb{R}$ to reduce to our example. Then $T^*G \cong \mathbb{R}^2$, $\mathfrak{g}^* \cong \mathbb{R}$, and the identity element of the group $\mathbb{R}$ is zero.

How does the momentum mapping $P: \mathbb{R}^2 \to \mathbb{R}$ coincide with the right translation of covectors to zero?

So I'm asking what exactly the map $P$ is and what Arnol'd and Givental mean by "the right translation of covectors to zero".