The lemma and its proof are given below:
My questions are:
1- In the second and third line of the proof, why is the case $i = j$ is not considered?
2- why we switch $i$ and $j$ in the second sum? and why this leads to that the second sum becomes the negative of the first?
You have to compute $\partial_{n-1}(\sigma \mid [v_0,\ldots,v_{i-1},v_{i+1},\ldots,v_n])$. Write $[v_0,\ldots,v_{i-1},v_{i+1},\ldots,v_n] =[w_0,\dots,w_{n-1}]$ with $w_j = v_j$ for $0< j < i$ and $w_j = v_{j+1}$ for $i \le j \le n-1$ shows that $$\partial_{n-1}(\sigma \mid [v_0,\ldots,v_{i-1},v_{i+1},\ldots,v_n]) = \partial_{n-1}(\sigma \mid [w_0,\ldots,w_{n-1}]) = \sum_{j=0}^{n-1}(-1)^j\sigma \mid [w_0,\ldots,\hat w_j,\ldots,w_{n-1}]) \\ = \sum_{j=0}^{i-1}(-1)^j\sigma \mid [w_0,\ldots,\hat w_j,\ldots,w_{n-1}]) + \sum_{j=i}^{n-1}(-1)^j\sigma \mid [w_0,\ldots,\hat w_j,\ldots,w_{n-1}]) \\ = \sum_{j=0}^{i-1}(-1)^j\sigma \mid [v_0,\ldots,\hat v_j,\ldots,\hat v_i,\ldots,v_n]) + \sum_{j=i}^{n-1}(-1)^j\sigma \mid [v_0,\ldots,\hat v_i,\ldots,\hat v_{j+1},\ldots,v_n]) \\ = \sum_{j=0}^{i-1}(-1)^j\sigma \mid [v_0,\ldots,\hat v_j,\ldots,\hat v_i,\ldots,v_n]) + \sum_{k=i+1}^{n}(-1)^{k-1}\sigma \mid [v_0,\ldots,\hat v_i,\ldots,\hat v_k,\ldots,v_n])$$ In the last line we replaced $j$ by $k-1$ so that $k = j+1$ and the summation goes from $k = i+1$ to $k=n-1+1 =n$. In the line before note that omitting $w_j$ for $j \ge i$ means omitting $v_{j+1}$.