In vacuum we have $R_{\mu\nu}=0$ and therefore $C_{\alpha\beta\mu\nu}=R_{\alpha\beta\mu\nu}$ therefore 
\begin{align}
\Psi_4
&=C_{\alpha\beta\mu\nu}\tilde{k}^\alpha \bar{\tilde{m}}^\beta \tilde{k}^\mu \bar{\tilde{m}}^\nu\\
&=R_{\alpha\beta\mu\nu}\tilde{k}^\alpha \bar{\tilde{m}}^\beta \tilde{k}^\mu \bar{\tilde{m}}^\nu\\
&=\frac{2}{i}R_{\alpha\beta\mu\nu}k^\alpha m^\beta k^\mu m^\nu
\end{align}
using the (rescaled) tetrad vectors
\begin{align}
\mathbf{k}&=\frac{1}{2}(\mathbf{e}_t-\mathbf{e}_r)\\
\tilde{\mathbf{k}}&=\frac{1}{\sqrt{2}}(\mathbf{e}_t-\mathbf{e}_r)\\
\mathbf{m}&=\frac{1}{2}\left(\mathbf{e}_\theta+\mathbf{e}_\phi+i(\mathbf{e}_\theta-\mathbf{e}_\phi)\right)\\
\tilde{\mathbf{m}}&=\frac{1}{\sqrt{2}}(\mathbf{e}_\theta+i\mathbf{e}_\phi)\\
\tilde{\bar{\mathbf{m}}}&=\frac{1-i}{\sqrt{2}}\mathbf{m}
\end{align}

Relevant components of the Riemann tensor for the linearised wave
\begin{align}
R_{TxTx}&=-\frac{1}{2}\partial_T^2h_+\\
R_{TyTy}&=+\frac{1}{2}\partial_T^2h_+\\
R_{TxTy}&=-\frac{1}{2}\partial_T^2h_\times
\end{align}
Then we just calculate and hope - thankfully the sixty terms reduce to
\begin{align}
R_{\alpha\beta\mu\nu}k^\alpha m^\beta k^\mu m^\nu
&=R_{TxTy}+\frac{i}{2}(R_{TxTx}-R_{TyTy})\\
&=-\frac{1}{2}\partial_T^2h_\times-\frac{i}{2}\partial_T^2h_+
\end{align}
and obtain
\begin{align}
\Psi_4
&=\frac{2}{i}\left(-\frac{1}{2}\partial_T^2h_\times-\frac{i}{2}\partial_T^2h_+\right)\\
&=-\partial_T^2h_+ +i\partial_T^2h_\times
\end{align}