Background Let $\mathcal H$ be a complex Hilbert space. Let $\langle \cdot,\cdot\rangle$ the scalar product of $\mathcal H$ (which is linear in the second argument) and $|\cdot|=\sqrt{\langle \cdot,\cdot\rangle}$. Let $\mathcal S\mathcal H=\{\psi\in \mathcal H\mid |\psi|=1\}$ be the Hilbert sphere.
For any $\phi\in \mathcal S\mathcal H$, $\{(\mathcal S\mathcal H\setminus \{\phi\},v_\phi),(\mathcal S\mathcal H\setminus \{-\phi\},v_{-\phi})\}$ is a Hilbert atlas on $\mathcal S\mathcal H$. \begin{align*} v_\phi&\colon \chi\in \mathcal S\mathcal H\setminus \{\phi\}\mapsto \frac{\chi-\langle \phi,\chi\rangle\phi}{1-\langle \phi,\chi\rangle}\in \phi^\perp \\ v_{-\phi}&\colon \chi\in \mathcal S\mathcal H\setminus \{-\phi\}\mapsto \frac{\chi-\langle \phi,\chi\rangle\phi}{1+\langle \phi,\chi\rangle}\in \phi^\perp \end{align*}
The collection $((U_\phi,u_\phi))_{\phi\in \mathcal S\mathcal H}$ is an atlas of the projective Hilbert space $\mathbb P \mathcal H$, where $U_\phi=\mathcal S\mathcal H\setminus \phi^\perp$ and $$ u_\phi\colon \chi\in U_\phi\mapsto \frac{\chi}{\langle \phi,\chi\rangle}-\phi\in \phi^\perp\,. $$
Question Now I want to find a suitable local description of the projection $\Pi\colon \mathcal S\mathcal H\to \mathbb P\mathcal H$. To do this, I fix $\psi\in \mathcal S\mathcal H$ and $\phi\in \psi^\perp$. Since $\psi\in \mathcal S\mathcal H\setminus \{\phi\}$, I consider the charts $(\mathcal S\mathcal H\setminus \{\phi\},v_\phi)$ and $(U_\psi,v_\psi)$, so that the local representation is $$ u_\psi\circ \Pi\circ v_\phi^{-1}\colon \phi^\perp\cap U_\psi\to \psi^\perp $$ Since $$ v_\phi^{-1}(\xi)=\frac{|\xi|^2-1}{1+|\xi|^2}\phi+\frac{2}{1+|\xi|^2}\xi\,, $$ I obtain $$ (u_\psi\circ \Pi\circ v_\phi^{-1})\xi=\frac{(|\xi|^2-1)\phi+2\xi}{2\langle \psi,\xi\rangle}-\psi=\frac{|\xi|^2-1}{2\langle\psi,\xi\rangle}\phi+\bigg(\frac{\xi}{\langle \psi,\xi\rangle}-\psi\bigg) $$
Are these calculations ok?
I'm editing my first answer because it needs to be corrected.
First, let me say that the expression of the chart $v_\phi$ is incorrect. Indeed, if $\dim_\mathbb C\mathcal H=N$ then $\mathcal S\mathcal H$ is a $(2N-1)$-real dimensional manifold. In particular, $\mathcal S\mathcal H$ can't be modelled on a $2(N-1)$-real dimensional vector space as $\phi^\perp$. The correct chart is $$ v_\phi\colon \chi\in \mathcal S\mathcal H\setminus \{\phi\}\mapsto \frac{\chi-\Re\langle \phi,\chi\rangle\psi}{1-\Re\langle \phi,\chi\rangle}\in \{\xi\in \mathcal H\mid \Re\langle \phi,\xi\rangle=0\}\,. $$ Fortunately, the $v_\phi^{-1}$ expression is correct.
Moreover, the correct expressions of $U_\phi$ and $u_\phi$ are $$ U_\psi=\Pi(\mathcal S\mathcal H\setminus \psi^\perp) $$ and $$ u_\phi\colon [\chi]\in U_\phi\mapsto \frac{\chi}{\langle \phi,\chi\rangle}-\phi\in \phi^\perp\,. $$
Finally, the locally representation of $\Pi$ at $\psi$ in charts $(\mathcal S\mathcal H\setminus \{-\psi\},v_{-\psi})$ and $(U_\psi,u_\psi)$ is $$ u_\psi\circ \Pi\circ v_{-\psi}^{-1}\colon v_\psi(\Pi^{-1}(U_\psi)\cap (\mathcal S\mathcal H\setminus \{-\psi\}))\mapsto u_\psi(U_\psi\cap\Pi(\mathcal S\mathcal H\setminus\{-\psi\})) $$
Let us observe that \begin{align*} v_\psi(\Pi^{-1}(U_\psi)\cap(\mathcal S\mathcal H\setminus \{-\psi\}))&=v_\psi((\mathcal S\mathcal H\setminus \psi^\perp)\cap(\mathcal S\mathcal H\setminus \{-\psi\}))\\ &=v_\psi(\mathcal S\mathcal H\setminus (\{-\psi\}\cup\psi^\perp))\\ &=\{\xi\in \mathcal H \mid \Re\langle \psi,\xi\rangle=0, |\xi|\ne 1\} \end{align*} and \begin{align*} u_\psi(U_\psi\cap \Pi(\mathcal S\mathcal H\setminus \{-\psi\})&=u_\psi(\Pi(\mathcal S\mathcal H\setminus \psi^\perp)\cap \Pi(\mathcal S\mathcal H\setminus \{-\psi\}))\\ &=u_\psi(U_\psi)\\ &=\psi^\perp\,. \end{align*}
Finally, \begin{align*} (u_\psi\circ \Pi\circ (\varphi_\psi^-)^{-1})(\xi)&=(u_\psi\circ \Pi)\bigg(-\frac{|\xi|^2-1}{1+|\xi|^2}\psi+\frac{2\xi}{1+|\xi|^2}\bigg)\\ &=u_\psi((1-|\xi|^2)\psi+2\xi)\\ &=\frac{(1-|\xi|^2)\psi+2\xi}{\langle \psi,(1-|\xi|^2)\psi+2\xi\rangle}-\psi\\ &=\frac{(1-|\xi|^2)\psi+2\xi}{1-|\xi|^2+2\langle \psi,\xi\rangle}-\psi\\ &=\frac{2(\xi-\langle \psi,\xi\rangle\psi)}{1-|\xi|^2+2\langle \psi,\xi\rangle} \end{align*}