Intuition behind a particular formulation of Bayes's Theorem : $\dfrac{P(A\mid B)}{P(A)} = \dfrac{P(B\mid A)}{P(B)}$?

I agree that this is less obvious, but you can go some way towards making intuitive sense of it by noting that it's obvious in three important cases:

If $A$ and $B$ are identical, it's obviously true by symmetry.

If $A$ and $B$ are independent, it's obviously true because both ratios must then be $1$.

If $A$ and $B$ are mutually exclusive, it's obviously true because both ratios must then be $0$.

Given this, it would be surprising if such simple ratios would manage to coincide at three different points but not in general.

You may be confusing yourself because you are skipping a step. Consider the definition of conditional probability. $$P(A|B)=\frac{P(A\cap B)}{P(B)}\implies P(A\cap B)=P(A|B)P(B)$$

The intersection of two sets is commutative, i.e. $$P(A\cap B) = P(B\cap A)=P(B|A)P(A)$$

Therefore, $P(A|B)P(B)=P(B|A)P(A)$