I would like to better understand the characterization of quasi-coherent sheaves on $X=\mathbb{P}_A^d = \operatorname{Proj}A[x_0,...x_d]$, where I call $B = A[x_0,...x_d]$
We define $\mathcal{F}[k] = \mathcal{F}\otimes_{\mathcal{O}}\mathcal{O}[k]$ for every quasi-coherent sheaf $\mathcal{F}$ on $X$, where $\mathcal{O}[n]$ is defined as $\mathcal{O}[n](D_+(f))=\mathcal{O}(X)[n]_{(f)}$.
claim: $$\mathcal{F} \cong \widetilde{\sum\mathcal{F}[k](X)}$$
proof:
Fact 1: to prove an isomorphism of sheaves, it suffices to prove that they agree on a basis of open sets, because this gives an isomorphism at the level of stalks, and the isomorphism at the level of stalks then reflects to an isomorphism of sheaves.
Hence, we are left to show that we have an isomorphism on each $D_+(x_i)$, as these provide a convenient base. The reason why these base is convenient is that $x_i$ is homogenous of degree $1$.
Fact 2: Since $x_i$ has degree 1, it follows that $\mathcal{O}[k](X)_{(x_i)} \cong x_i^k \mathcal{O}(X)_{(x_i)}$.
Fact 3: we have $\widetilde{\sum\mathcal{F}[k](X)}(D_+(x_i))=\left(\sum\mathcal{F}[k](X)\right)_{(x_i)} \cong \sum \mathcal{F}(X)\otimes \mathcal{O}(X)[k]_{(x_i)} \cong \sum \mathcal{F}(X)\otimes x_i^k \mathcal{O}(X)_{(x_i)}$.
We are then left to find an isomorphism $$\mathcal{F}(D_+(x_i)) \cong \sum_{\mathbb{Z}} \mathcal{F}(X)\otimes_{\mathcal{O}(X)} x_i^k \mathcal{O}(X)_{(x_i)}=M_{(x_i)}$$
Fact 5: An element $s \in M_{(x_i)}$ is, by definition, of type $s=v/x_i^k$ with $v$ homogenous of degree $k$, hence $v \in \mathcal{F}[k](X)$. Restricting to $D_{+}(x_i)$, we obtain, by distributivity, $v_{|D_{+}(x_i)}=g\otimes x_i^k \in \mathcal{F}(D_{+}(x_i))\otimes x_i^k\mathcal{O}(X)_{(x_i)}$ for some uniquely determined $g \in \mathcal{F}(D_{+}(x_i))$.
We define a map $M_{(x_i)}\to \mathcal{F}(D_{+}(x_i))$ given by $ s = v/x_i^k \mapsto v_{|D_{+}(x_i)}= g \otimes x_i^k \mapsto g$. Claim: this map is an isomorphism.
Fact 6: As $X=\mathbb{P}_A^d$ is compact and separated, for every invertible sheaf $\mathcal{L}$ and every section $l \in \mathcal{L}(X)$, we have that for every $g \in \mathcal{F}(X_l)$ there exists an $n_0 \geq 1$ such that $g \otimes (l_{|X_l}^{\otimes n})= v_{|X_l}$ for some $v \in (\mathcal{F}\otimes \mathcal{L}^{\otimes n})(X)$ for all $n \geq n_0$.
Surjectivity: taking $\mathcal{L}= \mathcal{O}$, and $l=x_i$ and $X_l=D_+(x_i)$, we obtain that for every $g \in \mathcal{F}(D_{+}(x_i))$ there exists a $n_0 \geq 1$ such that for all $n \geq n_0$ there exists a $v \in \mathcal{F}(X)\otimes_{\mathcal{O}(X)} \mathcal{O}(X)_n$ such that $v_{|D_+(x_i)}=g\otimes x_i^n$ and then one sends $g \otimes x_i^n \mapsto g$.
Fact 7: given $f \in \mathcal{F}(X)$ such that $f_{|X_{l}}=0$ there exists $n \geq 1$ such that $f \otimes l^{\otimes n} = 0$ in $(\mathcal{F} \otimes \mathcal{L}^{\otimes n})(X)$
Injectivity: We have seen with the proof of surjectivity that any element $g$ in the image of the map has the property that $g \otimes x_i^n=v_{|D_+(x_i)}$ for some big enough $n$ and some $v \in \mathcal{F}(X)\otimes_{\mathcal{O}(X)} \mathcal{O}(X)_n$. If $g = 0$, it means that $v_{|D_{+}(x_i)}=g \otimes x_i^n = 0$, which implies that for some $m \geq 1$ we have $v \otimes x_i^m = 0\in \mathcal{F}(X)\otimes_{\mathcal{O}(X)} \mathcal{O}(X)_n$, which implies $v=0$ and hence $v/x_i^k=0$.
question: Looks good?