Let $f\in L^2(\mathbb{R}^2)$ we denote it´s Fourier transform by $\hat{f}(\xi)=\int_{\mathbb{R}^2}f(x)e^{-2\pi i\langle x,\xi\rangle}dx$, where $\langle \cdot,\cdot\rangle$ is the inner product on $\mathbb{R}^2$.
For $\psi\in L^2(\mathbb{R}^2)$ we define the discrete shearlets as $\psi_{j,k,m}(x)=2^{\frac{3}{4}j}\psi(S_kA_{2^j}x-m)$, where $j,k\in\mathbb{Z}$, $m\in\mathbb{Z}^2$, $S_k=\pmatrix{1&k\\0&1}$ and $A_{2^j}=\pmatrix{2^j&0\\0&2^{j/2}}$.
I´m now trying to compute the Fourier transform of $\psi_{j,k,m}$ depending on $\hat{\psi}$.
For convenience, I will define $\psi_{j,k} = \psi_{j,k,0}$ and first compute $\hat{\psi}_{j,k}$.
Let $\Phi = S_{k} A_{2^j}$. Notice that $\Phi$ is invertible with $\Phi^{-1} = \pmatrix{2^{-j}&-k2^{-\frac{j}{2}}\\0&2^{-j/2}}$. In particular, $\Phi(\mathbb{R}^2) = \mathbb{R}^2$. Hence, as noted in the comments, we can write
\begin{align*} \hat{\psi}_{j,k}(\xi) =& \int_{\mathbb{R}^2} \psi \circ \Phi (x) \exp(- 2\pi i \langle \xi, \Phi \circ \Phi^{-1} x \rangle) dx \\=& 2^{-\frac{3j}{2}} \int_{\mathbb{R}^2} \psi(x) \exp(- 2\pi i \langle \xi, \Phi^{-1} x \rangle) dx \end{align*} since if $D\Phi$ denotes the jacobian of $\Phi$, one can compute that $\det D\Phi = 2^{\frac{3j}{2}}$.
Hence, we conclude that $$\hat{\psi}_{j,k}(\xi) = 2^{-\frac{3j}{2}} \hat{\psi}((\Phi^{-1})^*\xi)$$ where simple linear algebra allows us to explicitly compute $$(\Phi^{-1})^* = \pmatrix{2^{-j}&0\\-k2^{-\frac{j}{2}}&2^{-j/2}}$$
It only remains to account for the translation. If $f \in L^2(\mathbb{R}^2)$ and $m \in \mathbb{R}^2$, let $(T_mf)(x) = f(x-m)$. Then \begin{align*} \widehat{T_mf}(\xi) &= \int_{\mathbb{R}^2} f(x-m) \exp(-2\pi i \langle\xi, x \rangle ) dx \\ &= \int_{\mathbb{R}^2} f(x) \exp(-2\pi i \langle\xi, x+m \rangle ) dx \\ &= \exp(-2 \pi i \langle \xi, m \rangle) \hat{f}(\xi) \end{align*}
Hence, since $\psi_{j,k,m} = T_m \psi_{j,k}$ we can conclude that $$\hat{\psi}_{j,k,m}(\xi) = \widehat{T_m \psi_{j,k}}(\xi) = \exp(-2 \pi i \langle \xi, m \rangle)2^{-\frac{3j}{2}} \hat{\psi}((\Phi^{-1})^*\xi)$$