Why are significant figure rules in Multiplication/Division different than in Addition/Subtraction?

The number of significant figures is a representation of the uncertainty of a number. $123.4$ has an uncertainty of $0.1$ since the first uncertain digit is usually included. So, your multiplication of $$12.3 \times 4.6$$ is better represented as $$(12.3 \pm 0.1) (4.6 \pm 0.1).$$ I'm going to use @barrycarter's trick of using scientific notation to better represent the values $$(1.23 \pm 0.01) \cdot 10^1 \times (4.6 \pm 0.1) \cdot 10^0.$$ $$(1.23 \pm 0.01) (4.6 \pm 0.1) \times 10^1.$$ From this we can see that the number with more significant digits has less relative uncertainty. Relative uncertainty is approximately the measured uncertainty (also called absolute uncertainty) divided by the value.

Multiplying out: $$((1.23)(4.6) + (1.23)(\pm 0.1) + (4.6)(\pm 0.01) + (\pm 0.01)(\pm 0.1)) \times 10^1.$$ $$(5.658 + (\pm 0.123) + (\pm 0.046) + (\pm 0.001)) \times 10^1.$$ The number of digits we can write down is determined by the total uncertainty, which here is dominated by $\pm 0.123$. So, our result is $$(5.658 \pm 0.123) \times 10^1.$$ $$(5.7 \pm 0.1) \times 10^1.$$ $$57 \pm 1$$ which results in $57$.

To summarize, in multiplication, the largest relative uncertainty dominates the uncertainty of the final answer. This translates into the the number with fewer significant digits determining the significant digits of the final answer. This contrasts with addition, where the largest absolute uncertainty determines the uncertainty of the answer.