Let $f$ be a unitary operator. $\chi_s$ is the stabilizer operator where $s\in \{0,1,2,3\}^n$ where $\chi_s=\bigotimes\limits_{i=1}^n\sigma^{s_i}$ where $s_i\in \{0,1,2,3\}$. So we can write $f=\sum\limits_{s\in \{0,1,2,3\}^n}\hat{f_s}\chi_s$ where $\hat{f_s}=\langle f,\chi_s\rangle$.
Now we define $F_{s;I}=\frac1{2^{|I|}}\operatorname{tr}(\chi_tF_{s;I})$ where $I\subset [n]$ and $\operatorname{supp}(s)\subset I$. Then we have to prove $$\frac1{2^{n-|I|}}\operatorname{tr}(\chi_tF_{s;I})=\hat f\!_{s\cup t}$$where $s\cup t$ denotes the concatenation $$[s\cup t]_j=\begin{cases}s_j & j\in I\\ t_j & j\notin I\end{cases}$$where $t\in \{0,1,2,3\}^{n-|I|}$ with $\operatorname{supp}(t)\subset I^c$.
The proof goes like this $$F_{s;I}=\frac1{2^{n-|I|}}\sum\limits_{\operatorname{supp}(t)\subset I^c }\operatorname{tr}(\chi_tF_{s;I})\chi_t=\frac1{2^n}\sum\limits_{\operatorname{supp}(t)\subset I^c} \operatorname{tr}(\chi_s\otimes \chi_tf)\chi_t$$
How do we get this line? The first equality is still understandable but the second one I am not able to derive. Can you show the calculation for the two equalities here?