In vacuum we have $R_{\mu\nu}=0$ and therefore $C_{\alpha\beta\mu\nu}=R_{\alpha\beta\mu\nu}$ and so \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}