The qc open subsets of $\mathrm{Spec}(A)$ have the form $\cup_i D(f_i)$ with finitely many $f_i \in A$. If we intersect two such sets, we optain $\cup_i D(f_i) \cap \cup_j D(g_j) = \cup_{ij} D(f_i g_j)$, which is a finite union of affine schemes, and therefore qc.
Just to clarify, the real subtle point here is the following. For a $k$-scheme $X$ let us denote by $|X|$ the underlying topological space. Then, the behind-the-scenes complication here is that $|X\times_k X|\ne |X|\times |X|$. In fact, there is a continuous surjection $|X\times_k X|\to |X|\times |X|$ which is, in general, not an isomorphism (exercise!).
So, if we have a map $\phi:G\times_k G\to G$ and we have $(g,h)\in |G|$ there is no way to make sense of $\phi(g,h)$--one would have to try and lift them to $|G\times_k G|$ and apply $\phi$ there, but this is not well-defined (exercise!).
To convince yourself of why this is not so weird, consider $G=\mathrm{GL}_2$. Note then that we have that $H=\mathrm{SL}_2$ is a closed subscheme of $G$. Let $\eta$ denote the genric point of $G$ and $\eta'$ the generic point of $H$. Then, we have that $k(\eta)=k(x,y,z,w)$ and $k(\eta')=\mathrm{Frac}(k[x,y,z,w]/(xy-zw-1))$. The maps $ \mathrm{Spec}(k(\eta))\hookrightarrow G$ and $\mathrm{Spec}(k(\eta'))\hookrightarrow G$ correspond to the matrices $\begin{pmatrix}x & y\\ z & w\end{pmatrix}$ interpreted in $k(\eta)$ and $k(\eta')$ respectively. How do you 'multiply' those?
The point though is that while $|G\times_k G|\ne |G|\times |G|$ for every $k$-scheme $S$ we have that $(G\times_k G)(S)=G(S)\times G(S)$. Thus, for every $k$-scheme $S$ the map $\phi:(G\times_k G)(S)\to G(S)$ is a map $G(S)\times G(S)\to G(S)$ which is actually the multiplication for a group structure on $G(S)$.
Best Answer
The ring of global sections of the disjoint union (coproduct) is the product of the rings of global sections of the individual schemes: $$\bigcup^{\text{disjoint}}_{1\le i\le n} \text{Spec } R_i=\text{Spec }( R_1\times R_2\times \cdots\times R_n).$$
To explain this in more detail: Let $e_j$ be the element of $\hat R:=R_1\times\cdots\times R_n$ with $1$ in the $j$th position and $0$ elsewhere. Then, if $i\ne j$, any prime ideal $\hat P$ in $\hat R$ contains $e_i e_j=0$, so it contains either $e_i$ or $e_j$. Doing this for all pairs $\{i,j\}$ shows you that $\hat P$ must contain all of the $e_i$s except one, so it is of the form $R_1\times\cdots\times R_{i-1}\times P_i\times R_{i+1}\times\cdots\times R_n$, for some prime ideal $P_i$ of $R_i$. This shows that there is a natural correspondence between the points of $\text{Spec } \hat R$ and those of the disjoint union. Now, let $f_1\in R_1$, $\dots$, $f_n\in R_n$. Then the distinguished open set $D_{(f_1,\ldots,f_n)}$ of $\text{Spec } \hat R$ satisfies \begin{eqnarray*} D_{(f_1,\ldots,f_n)}&=&\{\hat P \subseteq \hat R\mid \hat P \text{ prime}, (f_1,\ldots,f_n)\notin \hat P\}\\ &=& \bigcup_{1\le i\le n} \{R_1\times\cdots\times P_i\times \cdots\times R_n\mid P_i \text{ prime}, f_i\notin P_i\}. \end{eqnarray*} Since these sets are a basis for the topology, if you then set $$ U_i:=D_{(0,\ldots,0,1,0,\ldots,0)}, \qquad i=1, \dots, n, \ \ \ \text{1 is at the }i\text{th position} $$ this shows that $$ \text{Spec } \hat R = U_1\cup\cdots\cup U_n $$ is a partition of the topological space of $\text{Spec } \hat R$ into $n$ disjoint clopen sets, and that the subset topology on each $U_i$ is identical with that of $\text{Spec } R_i$. Therefore, $\text{Spec } \hat R$ is the coproduct of the $\text{Spec } R_i$s in the category of topological spaces.
There is an isomorphism of localizations $${\hat R}_{(f_1,\ldots,f_n)} \cong (R_1)_{f_1}\times\cdots\times (R_n)_{f_n} $$ for which the square $$ \begin{array}{ccc} {\hat R}_{(f_1,\ldots,f_n)}&\cong&(R_1)_{f_1}\times\cdots\times(R_n)_{f_n}\\ \downarrow&\ &\downarrow\\ {\hat R}_{(g_1,\ldots,g_n)}&\cong&(R_1)_{g_1}\times\cdots\times(R_n)_{g_n} \end{array}$$ $$ (g_1 \in \sqrt{(f_1)}, \ \dots, \ g_n \in \sqrt{(f_n)}) $$ commutes. Since the structure sheaf of an affine scheme is assembled from these localizations, this can be used to show that there is an isomorphism $$ {\cal O}_{\text{Spec } \hat R}(U)\cong \prod_i {\cal O}_{\text{Spec } R_i} (\alpha_i^{-1}(U)) $$ for all open sets $U\subseteq \text{Spec } \hat R$, where the map $\alpha_i: \text{Spec } R_i\to U_i$ is the map of topological spaces which sends $\text{Spec } R_i$ onto its homeomorphic image inside $\text{Spec } \hat R$. Also, if $V\subseteq U$, the diagram $$ \begin{array}{ccc} {\cal O}_{\text{Spec } \hat R}(U)&\cong&\prod_i {\cal O}_{\text{Spec } R_i} (\alpha_i^{-1}(U))\\ \downarrow{}&&\downarrow{}\\ {\cal O}_{\text{Spec } \hat R}(V)&\cong&\prod_i {\cal O}_{\text{Spec } R_i} (\alpha_i^{-1}(V)) \end{array} $$ commutes. This means that the stalk at each point in each $\text{Spec } R_i$ is isomorphic to the stalk at its image in $\text{Spec } \hat R$.
For each $i=1$, $\dots$, $n$, there is a morphism $\gamma_i: \text{Spec } R_i\to \text{Spec } \hat R$ which comes from the ring homomorphism projecting $\hat R$ onto $R_i$. To show that $\text{Spec } \hat R$ is the coproduct of the $\text{Spec } R_i$s in the category of schemes, you need to show that for any scheme $Z$ and morphisms $\psi_i: \text{Spec } R_i \to Z$, there is a unique morphism $\psi: \text{Spec } \hat R\to Z$ such that $\psi \circ \gamma_i=\psi_i$ for all $i$. Since $\text{Spec } \hat R$ is the disjoint union of the $\text{Spec } R_i$s as topological spaces, the action of $\psi$ on the topology is determined by letting it act as $\psi_i$ on the copy $U_i$ of $\text{Spec } R_i$ in $\text{Spec } \hat R$. Now for any open set $W$ in $Z$, the morphisms $\psi_i$ give you ring homomorphisms from ${\cal O}_Z(W)$ to ${\cal O}_{\text{Spec } R_i}(\psi_i^{-1}(W))$. Using the isomorphism between $\prod_i {\cal O}_{\text{Spec } R_i}(\psi_i^{-1}(W))$ and ${\cal O}_{\text{Spec } \hat R}(\psi^{-1}(W))$, these can be assembled into a ring homomorphism from ${\cal O}_Z(W)$ to ${\cal O}_{\text{Spec } \hat R}( \psi^{-1}(W))$. This produces the desired morphism $\psi$.