I think you are asking how large (in various ways) can the discontinuity set of a function be if we assume that the discontinuity set has Lebesgue measure zero. Since any $F_{\sigma}$ set can be the discontinuity set of some function (and no other sets can be the discontinuity set of a function), your question is equivalent to asking how large (in various ways) can an $F_{\sigma}$ Lebesgue measure zero set be.
First, it’s worth mentioning that any such set is somewhat small. Indeed, it is easy to show that any such set must be a first (Baire) category set—use the fact that any closed measure zero set is first category, in fact even nowhere dense. Moreover, any such set is strictly smaller than some sets that are simultaneously first category and measure zero—see this 1 May 2000 sci.math post; also relevant is my answer to Jordan measure zero discontinuities a necessary condition for integrability.
However, it’s also possible that such a set can be uncountable, even have cardinality continuum, even have positive Hausdorff dimension, even have Hausdorff dimension $1.$ Note that these properties (in this order) give strictly stronger notions of “large”. In fact, it’s even possible that such a set can have the strongest of these properties (namely, Hausdorff dimension $1)$ in every nonempty open interval. To see this, let $\{I_n: \; n=1,2, \ldots\}$ be an enumeration of the rational-endpoint open intervals in $\mathbb R$ and let $C_n$ be a Cantor set (or any closed nowhere dense set) of Hausdorff dimension equal to $1 - \frac{1}{n}$ contained in $I_n.$ Note that each $C_n$ has Lebesgue measure zero. Then $D = \cup_{n=1}^{\infty} C_n$ is an $F_{\sigma}$ measure zero subset of $\mathbb R$ whose intersection with any nonempty open subinterval of $\mathbb R$ has Hausdorff dimension $1.$ The reason we get “everywhere of Hausdorff dimension $1$” is that given any nonempty open interval $I,$ there exist infinitely many $n$ such that $I_n \subseteq I,$ and so for every $\epsilon > 0$ and for every nonempty open interval $I,$ the set $D \cap I$ has Hausdorff dimension greater than $1 - \epsilon,$ which implies that $D \cap I$ has Hausdorff dimension $1.$
Your question also seems to ask about the possible “largness” of some specific types of discontinuity sets. It’s not clear to me exactly what you are asking, but the following observations are probably relevant. First, for any function the set of points at which the left and right limits of that function both exist (either finitely or infinitely) and are different is countable. This has been known since at least as early as 1907, and a discussion of this result and some stronger versions of it is given in my answer to A search for theorems which appear to have very few, if any hypotheses. Second, for any function the set of points at which the function has an infinite limit is also countable, and in fact, is a much smaller type of set than is the case for countable sets in general, since every such set of points is actually scattered (see paper 1 or paper 2). Being scattered is a type of smallness that fits between the smallness of isolated sets (all isolated sets are scattered, but a convergent sequence with its limit point is a scattered set that isn't an isolated set) and the smallness of nowhere dense sets (every scattered set is nowhere dense, but a nonempty nowhere dense perfect set—the Cantor set, for instance—is a nowhere dense set that isn't scattered). One can view a scattered set as a super-small type of nowhere dense set: A nowhere dense set has the property that it's closure is not dense in any interval, whereas a scattered set has the property that it's closure is not dense in any interval or even in any Cantor set.
