Suppose $X_i$ 's are i.i.d and by sum of random variables, then $Var(X_1 + X_2) = Var(X_1) + Var(X_2)$.
However, since $X_i$ are i.i.d, then $X_1 + X_2 = 2 X_1$, then $Var(X_1 + X_2) = Var(2X_1) = 4 Var(X_1)$.
Which part of my logic went wrong here ?
$X_1$ and $X_2$ are identically distributed random variables but this does not mean realisations of those random variables will result in identical numbers.
To make this more concrete, consider e.g. $$ X_1 \sim N(0,1) $$ $$ X_2 \sim N(0,1) $$ These are i.i.d. but it doesn't mean that $X_1+X_2 \sim 2\,N(0,1)$ as you suggest.