Intuition for expressing a cycle as a product of transpositions.I have recently learned the proof...

Step 1
The intuition you seem to need is how to read off from a permutation given as a product of cycles the mapping the permutation represents. If you consider a single cycle such as $(123)$, then it is the permutation that maps 1 to 2, 2 to 3 and 3 to 1. When you look at a product of two or more cycles that may overlap, then you have to decide whether to work from left to right or from right to left. If I work from left to right $(123)(24)$ maps 1 to 4 (via 2), 2 to 3, 3 to 1 and 4 to 2.
Step 2
So now it should be intuitively clear that $(12)(23)\dots (4748)(481)$ (for example) is the cyclic permutation $(12\dots 48)$. And, in general that if the $a_i$ are all distinct that $(a_1{a}_2)(a_2{a}_3)\dots (a_{n-1}{a}_n)(a_n{a}_1)$ is the cyclic permutation $(a_1{a}_2\dots a_n)$.

Step 1
$(a_1,a_2,\cdots ,a_n)=(a_1,a_n)(a_1,a_{n-1})\cdots (a_2,a_1)$
Read the composition of cycles from right to left. How do they act on the element $a_1$? The right most transposition "sees" $a_1$ and takes it to $a_2.$. None of the rest of the transpositions act on $a_2.$.
Consider the element $a_k,$, the first several transpositions do not act on $a_k.$. The first transposition that does is $(a_1,a_k).$. That takes $a_k$ to $a_1.$. The next transpositions takes that element to $a_{k+1}.$.
Step 2
In this chain of transpositions, $a_1$ is acting as the interchange for all of the other elements. Every element gets sent to $a_1$ and then sent from $a_1$ to its destination.
$(a_n,a_{n-1})\cdots (a_n,a_1)$ is the same idea, except $a_k$ is acting as the interchange.
$(a_1,a_2)(a_2,a_3)\cdots (a_{n-1},a_n)$ even though it is ultimately the same cycle, the activity that is going on is a little bit different. Consider the element $a_k,$, it glides along unaffected by the first several transpositions, until it hits $(a_{k+1},a_k)$ gets shifted and then again the rest of the transpositions do nothing.