Consider a compositional inverse pair of functions, $h$ and $h^{-1}$, analytic at the origin with $h(0)=0=h^{-1}(0)$.
Then with $\omega=h(z)$ and $g(z)=1/[dh(z)/dz]$,
$$\exp \left[ {t \cdot g(z)\frac{d}{{dz}}} \right]f(z) = \exp \left[ {t\frac{d}{{d\omega }}} \right]f[{h^{ - 1}}(\omega )] = f[{h^{ - 1}}[t + h(z)]] = f[L(t,z)]$$
(see OEIS A145271 and A139605),
so with $D_{FT}(\alpha)$ the Fourier transform of $D(z)$, formally
$$D\left( {t \cdot g(z)\frac{d}{{dz}}} \right)f(z) = \int\limits_{ - \infty }^\infty {{D_{FT}}} (\alpha )\exp \left[ {2\pi i\alpha t\cdot g(z)\frac{d}{{dz}}} \right]d\alpha f(z)$$
$$ = \int\limits_{ - \infty }^\infty {{D_{FT}}} (\alpha )f\left\{ {{h^{ - 1}}\left[ {2\pi i\alpha t + h(z)} \right]} \right\}d\alpha = \int\limits_{ - \infty }^\infty {{D_{FT}}} (\alpha )f\left[ {L\left( {2\pi i\alpha t,z} \right)} \right]d\alpha $$
For the special case $D(z)=\sin(2\pi a z)$, $D_{FT}=\dfrac{\delta(\alpha-a)- \delta(\alpha+a)}{2i}$,
and so
$$sin\left( {2\pi a \cdot g(z)\frac{d}{{dz}}} \right)f(z) = \frac{{f\left\{ {{h^{ - 1}}\left[ {h(z) + 2\pi ia} \right]} \right\} - f\left\{ {{h^{ - 1}}\left[ {h(z) - 2\pi ia} \right]} \right\}}}{{2i}}$$
(For a consistency check, try $h(z)=z$.)
Similarly, switch to the inverse Laplace transform to obtain formally
$$D\left( {t \cdot g(z)\frac{d}{{dz}}} \right)f(z) = \frac{1}{{2\pi i}}\int\limits_{\sigma - i\infty }^{\sigma + i\infty } {{D_{LPT}}} (p)\exp \left[ {pt \cdot g(z)\frac{d}{{dz}}} \right]dpf(z)$$
$$ = \frac{1}{{2\pi i}}\int\limits_{\sigma - i\infty }^{\sigma + i\infty } {{D_{LPT}}} \left( p \right)f\left\{ {{h^{ - 1}}\left[ {pt + h(z)} \right]} \right\}dp = \frac{1}{{2\pi i}}\int\limits_{\sigma - i\infty }^{\sigma + i\infty } {{D_{LPT}}} \left( p \right)f\left[ {L\left( {pt,z} \right)} \right]dp$$
For the special case $D(z)=\cosh(az)$, ${{\text{D}}_{LPT}}{\text{ = }}\frac{1}{2}\left[ {\frac{1}{{p - a}}{\text{ + }}\frac{1}{{p + a}}} \right]$,
and purely formally
$${\text{cosh}}\left[ {ag(z)\frac{d}{{dz}}} \right]f(z) = \frac{1}{{2\pi i}}\int\limits_{\sigma - i\infty }^{\sigma + i\infty } {\frac{1}{2}} \left[ {\frac{1}{{p - a}} + \frac{1}{{p + a}}} \right]f\left\{ {{h^{ - 1}}\left[ {p + h(z)} \right]} \right\}dp$$
$=\frac{1}{2}[f[h^{-1}[a+h(z)]]+ f[h^{-1}[-a+h(z)]]$.
Examples can be constructed from
$g(z)=(1+z)^{m+1}$, $h^{-1}(z)=(1-mz)^{-1/m}-1$, $h(z) = - \dfrac{{{{(1 + z)}^{ - m}} - 1}}{m}$, and
$L(t,z)=h^{-1}[t+h(z)]=[(1+z)^{-m}-mt]^{-1/m}-1$
with the limiting case for $m=0$ being
$g(z)=(1+z)$, $h^{-1}(z)= \exp(z)-1$, $h(z)= \log(1+z) $, and
$L(t,z)=h^{-1}[t+h(z)]=(1+z)e^{t}-1$.
Note for the Witt algebra that the actions are given by
$exp[tz^{m+1}d/dz]f(z)=f[z(1-mtz^{m})^{-1/m}]$.