Why are these two inequalities not the same even though they use the same equation?

The book used the second inequality, and gets a tigher result (stronger bounds) than you. Why?

This is because, using the first expression, you bound separately numerator and denominator, and that's not the optimal thing to do. Namely, you known $m \leq u_n \leq M$, and then bound $$ \frac{8 m-8}{M+2} \leq \frac{8 u_{n}-8}{u_n+2} \leq \frac{8 M-8}{m+2} \tag{1} $$ This is correct, but not optimal. You can't see immediately it from the expression you use, but using the second expression is equivalent to bounding it as $$ \frac{8 m-8}{m+2} \leq \frac{8 u_{n}-8}{u_n+2} \leq \frac{8 M-8}{M+2} \tag{2} $$ which is better (but, again, it is not not obvious you can do this based on the first expression$^{\dagger}$).

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${}^{(\dagger)}$ a way to realize you can indeed do this without using the second expression (as the book did) is to notice that the function $f\colon x\to \frac{8x-8}{x+2}$ is increasing, and so $f(m)\leq f(u_n) \leq f(M)$ whenever $m\leq u_n\leq M$.

I think I see what is going on. Let's focus on your left inequality. You basically do this: $$ U_{n+1} = \frac{8U_n-8}{U_n+2} \ge \frac{8\cdot3-8}{U_n+2}\ge \frac{8\cdot3-8}{4+2} = \frac{16}{6} $$ This is absolutely correct. However, we lose some tightness in the bound, because $\frac{8U_n-8}{U_n+2}$ is in fact increasing in $U_n$ (for $U_n>-2$). This is not obvious, but can be seen immediately in the rewritten form $8-\frac{24}{U_n+2}$. Therefore, you can replace $U_n$ with $3$ everywhere to get: $$ U_{n+1} = \frac{8U_n-8}{U_n+2} \ge \frac{8\cdot3-8}{3+2} = \frac{16}{5} $$