In this answer below (source), Why is it enough to show that $f_i|_{V_{ijk}}=f_j|_{V_{ijk}}$?
Show that the natural map $\alpha : \text{Hom}_{\operatorname{Sch}}(X,\text{Spec} A) \rightarrow \text{Hom}_{\operatorname{Ring}}(A,\Gamma(X,\mathcal{O}_X))$ is an isomorphism.
Let $g\in\hom_{ring}(A,\Gamma(X,\mathcal{O}_X)$.
Cover $X$ by affine open subsets $\{U_i=Spec(A_i)\}_{i\in I}$.
Now, the inclusion $U_i\hookrightarrow X$ gives us a map from global section of $U_i$ to global section of $X$ (i.e., $\rho^{X}_{Spec(A_i)}:\Gamma(X,\mathcal{O}_X)\rightarrow A_i$)
We take the composite map $A\hookrightarrow\Gamma(X,\mathcal{O}_X)\hookrightarrow A_i$
This gives rise to a map from $f_i:U_i=Spec(A_i)\rightarrow Spec(A)$ for each $i\in I$ (Note, $f_i$ is nothing but the Spec map of the composition of $g$ with the restriction map $\rho^{X}_{U_i}$, i.e., $f_i=Spec(\rho^{X}_{U_i}\circ g)$)
Notation: If $h:A\rightarrow B$ be a ring homomorphism, then the corresponding scheme morphism is denoted by $Spec(h):Spec(B)\rightarrow Spec(A)$
Now, we use that fact- If $X$ and $Y$ are two schemes, then giving a morphism from $X$ to $Y$ is equivalent to giving an open cover $\{U_i\}_{i\in I}$ of X, together with morphism $f_i:U_i\rightarrow Y$, where $U_i$ has the induced open subscheme structure, such that the restrictions of $f_i$ and $f_j$ to $U_i\cap U_j$ are the same, for each $i,j\in I$
Therefore, we need to check: $$ f_i|_{U_i\cap U_j}=f_j|_{U_i\cap U_j} $$ We need to cover $U_i\cap U_j$, again by affine open subsets (Otherwise, we cannot use the functoriality of $Spec$) Cover $U_i\cap U_j$ by $\{V_{ijk}=Spec(B_{ijk})\}_{k\in I}$
Enough to show, $f_i|_{V_{ijk}}=f_j|_{V_{ijk}}$ .
Why is it enough to show that $f_i|_{V_{ijk}}=f_j|_{V_{ijk}}$?
It seems circular to me. If we are taking another cover, we would have to show that the maps agree on pairwise intersections of this cover then.
