Identity rules can have different domains

Let $y=f(x)$, so $f^{-1}(y)=x$. From $y=2x+1$, rearrange to $x=(y-1)/2$, hence $f^{-1}(y)=(y-1)/2$. Rename the input to $x$. The original graph is an increasing line on $x>0$ with excluded endpoint $(0,1)$; therefore $R_f=(1,\infty)=D_{f^{-1}}$. The inverse definition includes this domain.

$f^{-1}(f(x))=((2x+1)-1)/2=x$. Its domain is the original $D_f=(0,\infty)$. Every intermediate output is in $D_{f^{-1}}=R_f$, so the composite exists on that entire domain.

$f(f^{-1}(x))=2((x-1)/2)+1=x$. Its domain is the original inverse domain $(1,\infty)$. The inverse outputs lie in $D_f$, so the composite exists on that entire domain.

The source comparison bubble answers the additional question explicitly: $D_{f^{-1}f}\ne D_{ff^{-1}}$. Both rules return their input, but one accepts every positive number while the other accepts only numbers greater than 1. For example, $1/2$ belongs only to the first domain. A rule alone does not define a function.