Why local boundedness gives joint continuity here and nothing like it holds in the real case

The locally bounded theorem on this page is not a disguised continuity theorem for arbitrary separate regularity. The false statement A bounded function of two real variables whose every coordinate slice is real analytic is continuous gives the concrete contrast: the real-valued function $f(x,y)=xy\mathrm{/}(x^2+y^2)$ off the origin, with $f(0,0)=0$, is bounded and every coordinate slice is real analytic, yet the function is not jointly continuous. The published B-page item named cex-partial-derivatives-without-continuity uses the same witness for a different purpose; it is named here only in prose, without a wikilink, because examples pages are leaves.

What changes in the holomorphic setting is not the bare word "separate" but the one-variable Cauchy theory. A bound on a holomorphic slice controls its derivative by Cauchy's inequalities bound every derivative by a boundary bound on a compactly contained circle, and the control is uniform in the remaining coordinates when the bound is uniform there. That is exactly the input used in A bounded separately holomorphic function on a polydisc is Lipschitz on every smaller polydisc: the bounded separately holomorphic function becomes locally Lipschitz, hence continuous, and then Osgood's lemma: continuous and separately holomorphic implies holomorphic upgrades that continuity plus separate holomorphy to full holomorphy.

So the hypothesis "locally bounded" is not decoration. It is the condition that turns one-variable holomorphic control into a joint estimate. The real-analytic counterexample shows that without the Cauchy inequality there is no reason for separate regularity and boundedness to force any joint continuity at all.