The math is sound. The method is not.
claim:
\[ \begin{aligned} & {\widehat{\left(\partial_x f\right)_n}}(k)=\frac{k}{2}\left(\hat{f}_{n-1}(k)-\hat{f}_{n+1}(k)\right) \\ & {\widehat{\left(\partial_y f\right)_n}}(k)=\frac{k}{2 i}\left(\hat{f}_{n-1}(k)+\hat{f}_{n+1}(k)\right) \end{aligned}\tag{1} \]
Setup
\[ \begin{aligned} & f(r, \theta)=\sum_{m \in \mathbb Z} f_m(r) e^{i m \theta} \\ & \hat{f_m}(k)=\int_0^{\infty} f_m(r) J_m(k r) r d r \end{aligned}\tag{2} \] Note that \((\partial_x f)_n\) involves treating \(\partial_x f\) as an independent function in its own right, rather than as the partial derivative of \(f\) and then applying the Azimuthal Fourier transform directly to this function to obtain the expansion: \[ \partial_x f=\sum_{m \in \mathbb Z}\left(\partial_x f\right)_m(r) e^{i m \theta}\tag{3} \]
or we may write it as
\[ \partial_x f(r, \theta)=\sum_{n \in \mathbb Z} g_n(r) e^{i n \theta}\tag{4} \]
From azimuthal expansion, we have
\[ g_n(r)=\frac{1}{2 \pi} \int_0^{2 \pi}(\partial x f(r, \theta)) e^{-i n \theta} d \theta\tag{5} \] As given before
\[ D_{ \pm}=\partial_{x} \pm i \partial_y=e^{ \pm i \theta}\left(\partial_r \pm \frac{i}{r} \partial_\theta\right)\tag{6} \]
acting on a single mode \(f_m(r) e^{i m \theta}\)
\[ \begin{aligned} &D_{+}\left[f_m e^{i m \theta}\right]=e^{i(m+1) \theta}\left(\partial_r-\frac{m}{r}\right) f_m\\ &D_{-}\left[f_m e^{i m \theta}\right]=e^{i(m-1) \theta}\left(\partial_r+\frac{m}{r}\right) f_m \end{aligned}\tag{7} \] As given before
\[ \partial_x f=\frac{1}{2}\left(D_{+}+D_{-}\right) f\tag{8} \]
it says
\[ \left(\partial_x f\right)_n=\frac{1}{2}\left[\left(D_{+} f\right)_n(r)+(D-f)_n(r)\right]\tag{9} \]
From (5), one obtains directly \[ \begin{aligned} \left(D_{+} f\right)_n(r) & =\frac{1}{2 \pi} \int_0^{2 \pi}\left(D_{+} f\right) e^{-i n \theta} d \theta \\ & =\frac{1}{2 \pi} \int_0^{2 \pi} D_{+}\left[\sum_m f_m e^{i m \theta}\right] e^{-i n \theta} d \theta \\ & =\frac{1}{2 \pi} \int_0^{2 \pi} \sum_m\left[D_{+} f_m e^{i m \theta}\right] e^{-i n \theta} d \theta \\ & =\frac{1}{2 \pi} \int_0^{2 \pi} \sum_m e^{i(m+1) \theta}\left(\partial_r-\frac{m}{r}\right) f_m e^{-i n \theta} d \theta \\ & =\sum_m\left(\partial_r-\frac{m}{r}\right) f_m \underbrace{\frac{1}{2 \pi} \int_0^{2 \pi} e^{i(m+1-n) \theta} d \theta}_{\delta_{m+1, n}} \\ & =\left(\partial_r-\frac{n-1}{r}\right) f_{n-1} \end{aligned}\tag{10} \] Similarly for \(D_{-}\)
\[ (D_{-}f)_n(r)=\left(\partial_r+\frac{n+1}{r}\right) f_{n+1}\tag{11} \]
Therefore
\[ \begin{aligned} & \left(\partial_x f\right)_n=\frac{1}{2}\left[\left(\partial_r-\frac{n-1}{r}\right) f_{n-1}+\left(\partial_r+\frac{n+1}{r}\right) f_{n+1}\right] \\ & \left(\partial_y f\right)_n=\frac{1}{2i}\left[\left(\partial_r-\frac{n-1}{r}\right) f_{n-1}-\left(\partial_r+\frac{n+1}{r}\right) f_{n+1}\right] \end{aligned}\tag{12} \]
Apply the order-\(n\) Hankel transform
\[ \begin{aligned} & \mathscr{H}_{n+1}\left[\left(\partial_r-\frac{n}{r}\right) f_n\right]=-k \hat{f}_n \\ & \mathscr{H}_{n-1}\left[\left(\partial_r+\frac{n}{r}\right) f_n\right]=k \hat{f}_n \end{aligned}\tag{13} \]
it says
\[ \begin{aligned} {\widehat{\left(\partial_x f\right)}}_n(k) & =\frac{1}{2}\left(-k \hat{f}_{n-1}+k \hat{f}_{n+1}\right) \\ & =\frac{k}{2}\left(\hat{f}_{n+1}-\hat{f}_{n-1}\right) \\ \left(\partial_{y} f\right)_n(k) & =\frac{k}{2 i}\left(\hat{f}_{n+1}-\hat{f}_{n-1}\right) \end{aligned}\tag{14} \]