Let us assume that $M$ is a compact Riemannian manifold (without boundary). For any point $x\in M$, we can pullback $C^\infty(M)$ functions to $T_x M$ via the exponential map, by setting $$ (\exp_x^* f)(X) = f(\exp_x(X)).$$ This will clearly be a bounded function, as $f$ itself must be bounded. However, in general, $\exp_x^* f$ will not have bounded derivatives, as the derivatives of the exponential function (hence the metric) enters the picture. In general, using ODE estimates for the Jacobi equation, one can get that $$ \bigl|D^k(\exp_x^* f)(X)\bigr| \leq \alpha_k\|f\|_{C^k(M)} e^{\beta_k |X|^2} $$ for some constants $\alpha_k, \beta_k \in \mathbb{R}$ that only depend on $M$, the metric and its derivatives.

My question is: Can we do better? For example, can we obtain $\beta_k=0$ with some additional curvature assumptions. Of course, there is the trivial flat case in which $\alpha_k=1$, $\beta_k=0$. But are there nontrivial statements one can make here?

Motivation: One could try to define a Fourier differential operator on $M$, by defining $$ (\mathscr{F}_\varepsilon u)(x, \xi) = \int_{T_xM} e^{-i\xi(X)} u(\exp_x(X)) \chi(\varepsilon|X|)d X,$$ where $\chi$ is a suitable cutoff function. For any $\varepsilon >0$, $u(\exp_x(X)) \chi(\varepsilon|X|)$ is a Schwartz function in fiber direction, hence its Fourier transform is well-defined. One could now try to let $\varepsilon \longrightarrow 0$ and ask if the limit exists. In the case that $\exp_x^*u$ and all derivatives are bounded, it does because then $\exp_x^*u \in \mathscr{S}^\prime(T_xM)$, hence it has a Fourier transform in $\mathscr{S}^\prime(T_xM)$. In the general case this probably doesn't work because the above estimate is not good enough. So I am asking if there are additional conditions that one could ask to make this work.