Let $U \le G$ be a subgroup of odd order of the finite group $G$. Suppose $t \notin U$ is an involution with $u^t \in uU'$ for all $u \in U$, where $U'$ denotes the commutator subgroup of $U$. Set $T := \langle t \rangle$. Let $\lambda \ne 1$ be a linear character of $U$ and extend it to linear character $\hat \lambda$ on $TU$ by setting $\hat \lambda(tu) = \hat \lambda(u) = \lambda(u)$ for all $u \in U$.
Set $S := TU \setminus T^U$. Then for $g \in G$ we have that $S^g \ne S$ implies $S^g \cap S = \emptyset$ and $N_G(S) = TU$, and the generalised character $1 - \hat \lambda$ is zero outside of $S$. Furthermore $$ ((1 - \hat\lambda), (1 - \hat\lambda))_{TU} = 2 $$ and $$ (1_{TU}, (1 - \hat \lambda))_{TU} = 1. $$
As $u^{-1}t = u^{-1}tx$ with $x \in U'$ and $\lambda(x) = 1$ we have $\hat \lambda(u^{-1} t u) = \hat \lambda(tu^{-1}xu) = \lambda(u^{-1}xu) = 1$, hence $1 - \hat \lambda$ is zero outside of $S$.
But why do we have $((1 - \hat\lambda), (1 - \hat\lambda))_{TU} = 2$ and $(1_{TU}, (1 - \hat \lambda))_{TU} = 1$?
Okay, this is indeed simple by using orthogonality relations and $\lambda \ne 1$ $$ ((1 - \hat\lambda), (1-\hat\lambda))_{TU} = (1,1)_{TU} - (1, \hat\lambda)_{TU} - (\hat\lambda, 1) + (\hat\lambda, \hat\lambda)_{TU} = 1 - 0 - 0 + 1 = 2 $$ and $$ (1_{TU}, (1 - \hat\lambda))_{TU} = (1_{TU}, 1_{TU})_{TU} - (1_{TU}, \hat\lambda)_{TU} = 1 - 0. $$