On Page 144 of "The Geometry of Schemes", we have the following:
$ Z_{i} = \text{Proj}\;k[x_{0},\dots,x_{n}]/(F_{i}) \subset \text{Proj}\;k[x_{0},\dots,x_{n}], $ so that $ \Gamma = \text{Proj}\;k[x_{0},\dots,x_{n}]/(F_{1},\dots,F_{r}). $
For any subset $ I = \lbrace i_{1},i_{2},\dots,i_{k} \rbrace \subset \lbrace 1,2,\dots,r \rbrace, $ we denote by $ |I| = k $ the number of elements of $ I, $ and by $ d_{I} = \sum_{\alpha = 1}^{k} d_{i_{\alpha}}, $ the sum of the degrees of the corresponding polynomials.
We then set $ M_{k} = \bigoplus_{|I|=k} S(-d_{I}) $ where $ S = k[x_{0},\dots,x_{n}]. $ Let $ M_{0} = S. $ We will write an element of $ M_{k} $ as a collection $ \lbrace G_{I} \rbrace $ of polynomials, where $ I $ ranges over all multi-indices of size $ k. $ By our definition, $ \lbrace G_{I} \rbrace $ will be homogeneous of degree $ d $ if $ \text{deg}(G_{I}) = d-d_{I} $ for each $ I. $
I think I'm fine up to here.
Define a complex
$$ 0 \rightarrow M_{r} \rightarrow M_{r-1} \rightarrow \dots \rightarrow M_{2} \rightarrow M_{1} \rightarrow M_{0} = S. $$
The map $ \varphi_{k}: M_{k} \rightarrow M_{k-1} $ is given by setting $ \varphi_{k}(\lbrace G_{I} \rbrace) $ equal to the collection of polynomials $ \lbrace H_{J} \rbrace, $ where $ H_{J} = \sum_{\alpha \notin J} \pm F_{\alpha}\cdot G_{J \cup \lbrace \alpha \rbrace} $ and the sign depends on the number of elements of $ J $ less than $ \alpha. $
How does the sign change depending on the number of elements of $ J $ less than $ \alpha?$
The image of $ \varphi_{1}:M_{1} \rightarrow M_{0} = S $ is exactly the ideal of $ \Gamma. $
I don't understand why this is true.
For your first question, the answer is the sign is $(-1)^n$, where $n$ is the number of elements of $J$ less than $\alpha$.
Edit:
The purpose of the sign is to make things cancel out so that the maps form a complex.
As an example, suppose we have $x_1$, $x_2$, and $x_3$ in $k[x_1,x_2,x_3]$ and consider the basis element $\{1,2,3\} $ of $M_3$. Applying $\phi_3$ we get $$x_1\{2,3\}-x_2\{1,3\}+x_3\{1,2\},$$ and then applying $\phi_2$, we get that the coefficient of $\{1\}$ is $$((-1)x_3(-x_2)+(-1)x_2(x_3))=x_2x_3-x_2x_3=0$$ for example.
In general, the idea is the following:
If we apply the map twice, then the coefficient of $I$ is $$\sum_{\alpha \not \in I} \sum_{\beta\ne \alpha \not\in I} (-1)^n(-1)^m(-1)^\epsilon F_\alpha F_\beta G_{I\cup\{\alpha,\beta\}},$$ where $n$ is the number of elements of $I$ less than $\alpha$, $m$ is the number of elements of $I$ less than $\beta$ and $\epsilon$ is 1 if $\beta > \alpha$ and $0$ otherwise. Note that if we collect the $G_{I\cup\{i,j\}}$ terms, we get one for $\alpha=i$, $\beta=j$ and one for $\alpha=j$, $\beta=i$. This $(-1)^\epsilon$ term is what makes these cancel out, since one of the terms will have $\epsilon=0$ and the other will have $\epsilon=1$.
End edit
For the second question, note that the elements of $M_1$ are $r$-tuples of polynomials, since there are $r$ one-element subsets of $\{1,\ldots,r\}$, and $$\phi(G_1,\ldots,G_r) = \sum_{\alpha=1}^r F_\alpha G_\alpha,$$ since at $M_0$, $J=\varnothing$, so every element of $\{1,\ldots,r\}$ are not contained in $J$, and they're all larger than $0$ elements of $J$. Then as $(G_1,\ldots,G_r)$ vary we get all the elements of $(F_1,\ldots,F_r)$.