This answer addresses the question "If $R$ is a commutative ring, is there any criterion for the property that any local section of the structure sheaf on $\text{Spec}(R)$ has open support?" As pointed out by Georges Elencwajg in the comments, if you're interested in the condition that any local section of any sheaf on $\text{Spec}(R)$ has open support, thinking about skyskraper sheaves will quickly show you that this happens if and only if $\text{Spec}(R)$ is finite and discrete.
First let me handle the case of global sections. To any $r\in R$, we can associate the ideal $I_r = \{s\in R\mid rs = 0\}$. If $p$ is a prime ideal, $r_p\neq 0$ if and only if $I_r\subseteq p$. So $\text{supp}(r) = V(I_r)$. Since $\text{supp}(r)$ is always closed, we're interested in the case when $\text{supp}(r)$ is clopen. But every clopen set in $\text{Spec}(R)$ has the form $V(e)$ for a unique idempotent element $e$ (see here). So $\text{supp}(r)$ is clopen if and only if $V(I_r) = V(e)$ if and only if $\sqrt{I_r} = \sqrt{(e)}$ for some idempotent $e$.
Now it's easy to translate this to the following condition. Let's call it $(\star)$:
For all $r$ there exists an idempotent $e$ such that $re = 0$ and for all $s$ such that $rs = 0$, some power of $s$ is a multiple of $e$.
We've shown that all supports of global sections are open if and only if $(\star)$.
It's not hard to check that $(\star)$ is closed under finite products and localization.
Now for local sections: If $s$ is a local section, then we can cover $\text{dom}(s)$ by basic affine opens $\bigcup_{i\in I}\text{Spec}(R_{f_i})$. Let $s_i$ be the global section of $\text{Spec}(R_{f_i})$ corresponding to $s$. Then the support of $s$ is open in $\text{dom}(s)$ if and only if the support of $s_i$ is open in $\text{Spec}(R_{f_i})$ for all $i\in I$. We conclude that all supports of local sections are open if and only if $(\star)$ holds in $R_f$ for every $f\in R$. But since $(\star)$ is closed under localization, we also get:
All supports of local sections are open if and only if $(\star)$.
Of course integral domains satisfy $(\star)$, taking $e = 1$ if $r = 0$ and $e = 0$ otherwise.
More generally, a ring is sometimes called primary if $(0)$ is a primary ideal. Equivalently, if every zero-divisor is nilpotent. Primary rings also satisfy $(\star)$, again taking $e=1$ if $r=0$ and $e=0$ otherwise. In fact, by the above argument you can see that $R$ is primary if and only if every global section $r$ has $\text{supp}(r) = \emptyset$ or $\text{supp}(r) = \text{Spec}(R)$.
So a sufficient condition for $(\star)$ is that $R$ is a finite product of primary rings. The condition in Keenan Kidwell's answer is a special case of this, since his rings have finite discrete $\text{Spec}(R)$, so they are finite products of rings with a unique prime ideal, which are necessarily primary.
Another sufficient condition for $(\star)$ is that $R$ is a Boolean ring, i.e. every element is idempotent (then the support of $r$ is $V(1-r)$, the complement of $V(r)$). This gives new examples, since no infinite Boolean ring is a finite product of primary rings (because primary rings have no idempotents other than $0$ and $1$).
Another way of looking at this is that every idempotent $e$ splits the ring $R$ into a product $R\cong R/(e)\times R/(1-e)$. So $(\star)$ has the following equivalent form: for all $r$ we can represent the ring as a product $R\cong R_1\times R_2$ (either of the factors may be the zero ring!) such that $\pi_2(r) = 0$ and $\pi_1(r) = r'$ has the property that if $r's = 0$, then $s$ is nilpotent in $R_1$. If $R$ is a finite product of primary rings, then this product decomposition will always just be a partition of the factors into two pieces. But in general, the decomposition as a product may depend heavily on the choice of $r$. For example, if $R$ is Boolean, the decomposition as a product is different for every $r$.
