Action of the diffeomorphism group on a tensor bundle I have two questions in two paragraphs I'm reading from the book Heat kernels and Dirac operators page 17.
$\textbf{ paragraph 1}$

Let $TM$ be the tangent bundle of $M$. A section $X \in \Gamma(M,TM)$ is called a vector field on $M$. If $\phi: M_1 \rightarrow M_2$ is a smooth map, it induces a map $\phi_*: TM_1 \rightarrow TM_2$, in such a way that $$\phi_* (v) \in T_{\phi(x)}M_2 \quad if \quad v \in T_xM_1. $$ In particular, diffeomorphisms $\phi$ of $M$ induce a diffeomorphism $\phi_*$ of $TM$. Informally, $TM$ is a Diff($M$)-equivariant bundle.

$\textbf{Question 1 :}$ I didn't get  what are the authors trying to say in the last sentence " Informally, $TM$ is a Diff($M$)-equivariant bundle" ?
$\textbf{Paragraph 2}$

Let $GL(M)= GL(TM)$ be the frame bundle obtained by applying the construction of proposition 1.4 to the tangent bundle $TM$; it has the structure group $GL(n)$ with $n= dim(M).$  From this principal bundle, we can construct a vector bundle on $M$ corresponding to any representation $E$ of $GL(n)$; these vector bundles are called tensor bundles. We see from this that any tensor bundle carries a natural action of Diff$(M)$, making it into a Diff$(M)$-equivariant bundle over $M$.

$\textbf{Question 2:}$ what is the natural action of Diff$(M)$ on a tensor bundle ?
 A: Regarding Question 1: if $\phi$ is a diffeomorphism $\phi\in\text{Diff}(M)$ then, as quoted, $\phi_*:TM\rightarrow TM$ is a vector bundle isomorphism, since by naturality of the tangent map it has an inverse given by $(\phi^{-1})_*$
In general, if you have a vector bundle $E$ over $M$, and you have an diffeomorphism $\phi$ of $M$, you can construct the pullback vector bundle $\phi^*E$, fitting the following commutative diagram
$$\begin{array}
*\phi^*E & \stackrel{\hat{\phi}}{\longrightarrow} & E \\
\downarrow{} & & \downarrow{} \\
X & \stackrel{\phi}{\longrightarrow} & X  
\end{array}
$$
It sometimes happens that the pullback bundle $\phi^*E$ is isomorphic to the original one; sometimes it is not. In our context, if you take $E=TM$, for any diffeomorphism $\phi$ of $M$, then $\phi^*TM$ will be isomorphic to $TM$, and the $\hat{\phi}$ in the above diagram will be precisely the tangent map $\phi_*$.
Regarding Question 2: as stated above, for vectors (elements of the tangent bundle), any diffeomorphism $\phi\in\text{Diff}(M)$ induces a natural map $\phi_*: TM\rightarrow TM$. For contravariant tensors, this action extends naturally, and the issue is really an issue of linear algebra.
Specifically, assume your $M$ is $n$-dimensional. A diffeomorphism acts on a contravariant tensor by "the tensorization of the representation on vectors", meaning the following: an element of the tensor bundle $T\in (TM)^{\otimes r}$ over a point $p\in M$ will be mapped to another tensor of this bundle over the point $\phi(p)\in M$.Since any tensor can be written as linear combination of basic elements, you can write
$$
T=\sum_{i\in I} \alpha_{i} v^1_{i}\otimes v^2_{i} \otimes \dots \otimes v^r_{i}
$$
where each $v^a_i$ is an element of $TM$, on the fiber over $p$. Then, the natural action of $\phi$ on this $T$ can be expressed by
$$
\phi_*T=\sum_{i\in I} \alpha_{i} (\phi_*v^1_{i})\otimes (\phi_*v^2_{i}) \otimes \dots \otimes (\phi_*v^r_{i})
$$
This works for contravariant tensors. If you have covariant tensors, i.e. elements or sections of $(T^*M)^{\otimes s}$ you must perform the pullback through $\phi^{-1}$. By the same argument as for the tangent bundle, the pullback construction maps the tensor bundle isomorphically into itself; one says that this bundles are $\text{Diff}(M)$-equivariant.
This statement is particularly clear if you identify (isomorphism classes) of vector bundles over $M$ in terms of their transition functions, the transition functions of the pullback under $\phi$ will be precisely $g_{\alpha\beta}\circ \phi: \phi^{-1}(U_\alpha\cap U_\beta)\rightarrow GL(n)$, where $\{U_\alpha,\psi_\alpha\}$ is a coordinate cover of $M$ and $g_{\alpha\beta}$ is the jacobian of $\psi_\alpha\circ\psi^{-1}_\beta$.
