The problem here does not lie in the mathematics but in my reading capapabilities. In fact, the correct Theorem 2.24 says

If $\chi \in \text{Irr}(B)$ is real-valued, then there are at most two irreducible Brauer characters which are constituents of $\widehat{\chi} = \chi\mid_{G_{p'}}$ and which are also real-valued.

For such a character $\varphi$ we have ${\varphi}^* = \varphi$ and this is exactly what is used in 6.

The problem here does not lie in the mathematics but in my reading capapabilities. In fact, the correct Theorem 2.24 says

If $\chi \in \text{Irr}(B)$ is real-valued, then there are at most two real-valued irreducible Brauer characters which are constituents of $\widehat{\chi} = \chi\mid_{G_{p'}}$.

For such a character $\varphi$ we have ${\varphi}^* = \varphi$ and this is exactly what is used in 6.

The problem here does not lie in the mathematics but in my reading capapabilities. In fact, the correct Theorem 2.24 says

If $\chi \in \text{Irr}(B)$ is real-valued, then $\widehat{\chi} = \chi\mid_{G_{p'}}$ is the sum of at most two irreducible Brauer characters that are also real-valued.

For such a character $\varphi$ we have ${\varphi}^* = \varphi$ and this is exactly what is used in 6.

The problem here does not lie in the mathematics but in my reading capapabilities. In fact, the correct Theorem 2.24 says

If $\chi \in \text{Irr}(B)$ is real-valued, then there are at most two irreducible Brauer characters which are constituents of $\widehat{\chi} = \chi\mid_{G_{p'}}$ and which are also real-valued.

For such a character $\varphi$ we have ${\varphi}^* = \varphi$ and this is exactly what is used in 6.