and Is removing a carboxyl group leads to an oxidation reaction? I feel like I need to understand how these terms are linked instead of just memorizing them. Could you explain to me how these terms are connected?
This is an excellent question with a short answer and a long answer. The short answer is that decarboxylation reactions
in which the carboxyl group is lost as CO2 involve oxidation. You can confirm this by calculating the oxidation states for carbon. In a carboxyl group, C has an oxidation state of +3. In CO2, C has an oxidation state of +4. The oxidation state increasing means, by definition, that the carbon was oxidized. If you search the scientific literature, you can find examples of non-oxidative decarboxylation, but they involve different processes.
On a broader note, this points to the importance of being able to calculate the oxidation state of carbon in various organic compounds. There are some solid tutorials online about how to do this, and it's a good skill to have, because the MCAT does expect you to understand the oxidation/reduction hierarchy of organic compounds and be able to recognize quickly, for example, that alcohols are reduced relative to ketones, or that alkenes are oxidized relative to alkanes. You're unlikely to encounter a question that straight-up asks you, say, what is the oxidation state of carbon in CH4, or CH3OH, or whatever (although it's not impossible!), but in principle you should be able to figure it out as a way to double-check your work if you come across some kind of biological redox reaction.
In terms of the terminology regarding pyruvate oxidation, one of the reasons why this term is used is that it aligns with the overall logic of carbohydrate (and fatty-acid) metabolism—which, in a nutshell, is that reduced biomolecules are oxidized to generate electrons, which are pushed to the ETC by the electron carriers FAD/FADH2 and NAD+/NADH. This, by the way, is a high-level concept that the AAMC emphasizes multiple times in the content outline, so it's good to see how every individual step fits into the picture.