In the end, the reason to choose things in that way is because it works, but in this case it's possible to explain what's going on.

For a warm-up, you can get some intuition for the method by looking at an easier English example of the same phenomenon:

", when preceded by itself in quotation marks, yields a true sentence.", when preceded by itself in quotation marks, yields a true sentence.

To explain the proof you gave, I will use $|n|$ to denote the formula what has Gödel number $n$. In the proof above, $H(n)$ should be read as asserting $F([|n|(n)])$, that is:

$H(n)$ says: $F$ holds of the number of the formula obtained by substituting $n$ into $|n|$

Thus $H([H])$ says,

$F$ holds of the number of the formula obtained by substituting $[H]$ into $|[H]|$

that is,

$F$ holds of the number of the formula obtained by substituting $[H]$ into $H$

So $H([H])$ asserts $F([H([H])])$.

(For informal clarity, I have intensionally not used underlines here.)

In the end, the reason to choose things in that way is because it works, but in this case it's possible to explain what's going on.

For a warm-up, you can get some intuition for the method by looking at an easier English example of the same phenomenon:

", when preceded by itself in quotation marks, yields a true sentence.", when preceded by itself in quotation marks, yields a true sentence.

To explain the proof you gave, I will use $|n|$ to denote the formula that has Gödel number $n$. In the proof above, $H(n)$ should be read as asserting $F([|n|(n)])$, that is:

$H(n)$ says: $F$ holds of the number of the formula obtained by substituting $n$ into $|n|$

Thus $H([H])$ says,

$F$ holds of the number of the formula obtained by substituting $[H]$ into $|[H]|$

that is,

$F$ holds of the number of the formula obtained by substituting $[H]$ into $H$

So $H([H])$ asserts $F([H([H])])$.

(For informal clarity, I have intensionally not used underlines here.)

Addendum. It's impossible in general to construct a formula $J$ such that $J$ is literally the same formula as $F([J])$. Any typical Gödel numbering has the property that $n < [F(\underline n)]$ for every formula $F(x)$ and number $n$. But if $J = F([\underline J])$ then $[J] = [F(\underline {[J]})]$.

So, if our proof method is going to work with an arbitrary Gödel numbering, is has to be more indirect. The proof gives a formula $H$ so that $H([H])$ is provably equivalent to $F([H([H])])$, but not literally the same formula. This may help explain why proof proceeds the way it does.