This is not much of answer, but I just want to state that in my opinion the simplest proof of the result is the one the OP has already alluded to:
1) Every convex function $\varphi$ is continuous.
2) If $f: [a,b] \rightarrow \mathbb{R}$ is Riemann integrable and $\varphi: \mathbb{R} \rightarrow \mathbb{R}$ is continuous, then $\varphi \circ f: [a,b] \rightarrow \mathbb{R}$ is Riemann integrable.
The proof of 1) can be found in most "honors calculus" texts (e.g. Spivak). If anyone wants a link to a proof, please let me know. The proof of 2) given in the (nicely written, very elementary, expository) article by Jitan Lu cited by the OP works from the "Riemann notion of Riemann integrability", i.e., convergence of the Riemann sums uniformly in the mesh of the partition. A lot of discussions of the Riemann integral (e.g. Spivak, Rudin) introduce the "Darboux notion of Riemann integrability" -- i.e., the one with the upper integral and the lower integral -- and with this notion the proof is shorter: closer to half a page than a full page. See for instance the proof given on page 7 of these notes, which is taken almost verbatim from Russell Gordon's analysis text.
I suppose it would be of some interest if there were an even shorter proof that took advantage of the convexity, but in practice if you are at all interested in theorems like this you are going to want to know that $\varphi \circ f$ is Riemann integrable even for not necessarily convex continuous $\varphi$.