Calcium ion and sulfate did not form precipitate
While calcium sulfate is usually termed insoluble, it is not the ‘sitting at the bottom like a rock’ type insoluble; rather, it is the ‘there’s no practical way for me to get the two ions into the same solution without precipitation, but I’m still able to identify both ions in solution’ type insoluble. Those sentences don’t really help, so let’s look at numbers.
\begin{array}{lcr}
\hline
\text{salt} &amp; \text{mass solubility} &amp; \text{molar solubility} \\ \hline
\ce{BaSO4} &amp; \pu{2.45e-3 g/l} &amp; \pu{1.05e-5 mol/l}\\
\ce{CaSO4} &amp; \pu{2.1 g/l} &amp; \pu{1.54e-2 mol/l}\\
\ce{CaC2O4} &amp; \pu{6.7e-4 g/l} &amp; \pu{5.23e-6 mol/l}\\
\hline
\end{array}
To calculate your actual experiment, you would have to transform these values into $K_\text{sp}$ values and then calculate how much would precipitate at each given step. Without wanting to do that tedious calculation, you can still arrive at the general conclusion:

addition of sulphate precipitates practically all barium ions

addition of sulphate precipitates a large part of calcium ions
but: a non-neglegible amount of $\ce{CaSO4}$ remains dissolved

addition of oxalate precipitates practically all the remaining calcium.


In fact, the solubilities of $\ce{BaSO4, SrSO4}$ and $\ce{CaSO4}$ are such that a saturated solution of $\ce{SrSO4}$ will always give precipitate when barium is added but never with calcium or strontium; while a saturated solution of $\ce{CaSO4}$ will give a precipitate if either strontium or barium is added but never with calcium. This can be used to identify the heaviest alkaline earth metal in an unknown sample.