A question about extending group actions to maps.

It is known that an action $\alpha: G \times M \rightarrow M$ induces for each group element $g \in G$ a transformation on $M$, $T_g : M \rightarrow M.$

Now what is done to extend this action to $\mathcal{F}(M):= \{f : M \rightarrow \mathbb{R}\}$ is to define a transformation $\mathbb{T}_g : \mathcal{F}(M) \rightarrow \mathcal{F}(M)$ by $\mathbb{T}_gf = f'$ with

$$f'(T_g(x)) = f(x) \, \,.$$

Why the above thing is defined in such a way? Also why do we define it to take as input $T_g(x)$?

I mean, of course we have that $T_g : M \rightarrow M$ is an automorphism, hence any $x \in M$ can be written as $x = T_g(x')$ for some $x' \in M$, but I simply don't understand why the above definition is formalized in that way.

Also, an equivalent (why?) definition, would be stating:

$$\mathbb{T}_gf(x') = f(T_{g^{-1}}(x')) \, \, \forall x' \in M .$$

This second one actually makes a bit more sense to me, in light of $\mathbb{T}_g$, but why do we take $g^{-1}$?

I will answer the second question first. If $G$ acts on $M$ from the left, then it acts on $\mathcal F(M)$ from the right. If you want to define an action from the left, i.e.,

$$\forall x \in M, \ \forall f \in \mathcal F(M), \ \forall g,h \in G \colon \quad (\mathbb T_{gh} f)(x) \stackrel{!}= (\mathbb T_g (\mathbb T_h f))(x).$$

Hence, if you "define" $(\mathbb T_gf)(x) = f(T_{g}(x))$, then the above equation reads

$$f(T_{gh}x) = f(T_h T_g x).$$

Since $T_{gh} = T_g T_h$, this is only correct if $G$ is Abelian. In other words: you get an action from the right. Sometimes one wants to write it as an action from the left, and thus one takes the inverse.

Now for the other questions. If $f'(T_g x) = f(x)$ and $f' = \mathbb T_g f$ by definition, then $f'(x) = f(T_{g^{-1}} x)$, which is the second version you stated.

As for the question why everything is defined as it is: well, because we get an action in this case.