Maybe I'm missing something, but it seems to me like the first question is just a simple definition chase. A morphism $\mathrm{Spec}(A)\to X$ is determined by finitely many affine open subsets $\mathrm{Spec}(B_n) \subseteq X$ and some ring maps $B_n\to A_{f_n}$ to localizations of $A$ satisfying certain compatibility properties. There are not so many elements of $A$ that are needed to witness all of this. Specifically, we need to witness the $f_n$, numerators of everything in the images of the maps $B_n\to A_{f_n}$, some elements annihilated by powers of the $f_n$ that witness that the maps $B_n\to A_{f_n}$ are actually homomorphisms and are appropriately compatible, and elements of $A$ that witness that some $f_n$ are in the ideal generated by others (and that $1$ is in the ideal generated by all of them). In total, it is clear that some subring $A_0\subseteq A$ of cardinality at most $\aleph_0+\sum |B_n|$ contains all the needed witnesses, and our morphism will factor through $\mathrm{Spec}(A_0)$.
For the second question, consider two factorizations $\mathrm{Spec}(A)\to\mathrm{Spec}(A_i)\to X$ ($i=0,1$) where $|A_i|<\kappa_0$ for some cardinal $\kappa_0$ depending only on $X$. Let $Y=\mathrm{Spec}(A_0)\times_X \mathrm{Spec}(A_1)$; this may not be affine, but it will still satisfy some cardinality bound depending on $X$. Applying the previous paragraph with $Y$ in place of $X$, we can factor the map $\mathrm{Spec}(A)\to Y$ through $\mathrm{Spec}(A_2)$, and we can bound $A_2$ by some cardinal $\kappa_1$ depending only on $X$. Now repeat this argument allowing $A_0$ and $A_1$ to have cardinality up to $\kappa_1$, and get some new bound $\kappa_2$ on the cardinality of $A_2$. Repeating this inductively, we get a sequence of cardinals $\kappa_n$; let $\kappa=\sup \kappa_n$. Then whenever $|A_0|<\kappa$ and $|A_1|<\kappa$, we can find $A_2$ with $|A_2|<\kappa$ that connects them. (Actually, I'm pretty sure that if you choose $\kappa_0$ in the obvious way, $\kappa_1$ will just be the same as $\kappa_0$, so none of this iteration is actually necessary.)