How do I calculate the actual yield? Isn't it what I "should have gotten" based on my experiment?

Yeah, you can say it that way (but then what is "actual"?). A way to further clarify these terms is to put things into context.

Suppose that you are performing the following reaction:

#"AlCl"_3(aq) + 3"H"_2"O"(l) -> "Al"("OH")_3(aq) + 3"HCl"(aq)#

(which is basically one of the steps to how aluminum can acidify water.)

Let's say you poured #"AlCl"_3(aq)# (#"50 mL"#, #"0.05 M"#), which is mostly dissolved in #"HCl"#, into water until you reach #"100 mL"#. The intention is to replace all three #"Cl"# atoms with the #"OH"# on water, bringing more #"HCl"# into solution. Let's say you wanted to figure out the mass of #"HCl"# produced.

From the concentration and volume you added, you can determine the number of #"mol"#s of #"HCl"# you have. From the molar mass, you can get the mass of #"HCl"#.

#\mathbf"Theoretical Yield"#

#= (0.05 cancel("mol AlCl"_3))/cancel("L") xx cancel("1 L")/(1000 cancel"mL") xx 50 cancel"mL" xx (3 cancel"mol HCl")/cancel("1 mol AlCl"_3) xx ("36.4609 g HCl")/cancel("mol HCl")#

#= "0.2735 g HCl"(aq)#

That's the yield you expect to get based on what you started with. Now, if you experimentally recovered #"0.2000 g HCl"#, then that would be your #\mathbf"actual yield"#.

So, "actual" then refers to your experimental results.

Lastly, given those two:

#\mathbf"% Yield"#

#= "Actual Yield"/"Theoretical Yield"xx100%#

#= 0.2000/0.2735xx100% = 73.14%#