This example is taken from here.
A quasigroup of order n is a pair (Q, o), where Q is a set of size n and "o" is a binary operation on Q s. t. for every pair of elements $a,b\in Q$ the equations $a$ o $x = b$ and $y$ o $a = b$ have unique solutions.
for example see the figure, I'm trying to figure out what is the operation. So far, I know it is closed under this operation because each value in the table is between 1-5. Honestly, the definition cannot tell how to compute the table.
Answer to your question: you're given multiplication table. So result of $1 \circ 3$ is $2$, and $2 \circ 4$ is $1$.
Long comment: there's a general notion of a magma $(X, *)$, which is just a set $X$ with an operation $*$, i. e. arbitrary map $$X \times X \rightarrow X, \quad (x_1, x_2) \mapsto x_1 * x_2 $$ (and your multiplication table is perfect example of such a map). But you can think about a map from $X \times X$ as bunch of maps from "right $X$" labeled by elements of "left $X$" — that is, you'll completely decribe your operation $*$ if you define individual maps of left multiplication by $x$: $(x\,* -)$ for each element $x \in X$; this approach will give you $$X \to Fun(X, X), \quad x \mapsto (x\,* -)$$ It's pretty obvious that every operation gives unique labeling of elements by their left multiplication maps and vise versa; same for right multiplication. But you need to be careful there: functions are composition-associative, whereas your operation is not guaranteed to be associative! So composing multiplication-by-$x$ with multiplication-by-$y$ (as maps from $X$ to itself) is not same as multiplying by $y * x$. (in your example $(1 \circ (1 \circ 2)) = 4 \neq 5 = ((1 \circ 1) \circ 2)$)
Now back to your example. Quasigroup is a particular type of a magma, where left and right multiplication operators are bijective (mentioned equations have unique solutions $\iff$ preimage of each element under every (left or right) multiplication-by-something is one element). So, quasigroup operation is a map $$X \times X \rightarrow X, \quad (x_1, x_2) \mapsto x_1 \circ x_2 $$ which restricts to a bijections $$\{x\} \times X \cong X, \, X \times \{x\} \cong X$$ for each $x \in X$. And it's exactly the definition of a latin square — every row $(\{x\} \times X)$ and every column $(X \times \{x\})$ contain each number exactly once (bijectivity).
Also note that not every map $$X \to Sym(X), \quad x \mapsto (x\,* -)$$ will give a quasigroup — for example, sending all $X$ to one element will not succeed; injectivity is not enough too, and you can construct counterexample very easy.