The functions $\xi(s)$ and $$f(s):=\xi\left(\frac{s-i}{2}\right) + \xi\left(\frac{i+s}{2}\right)$$ grow faster than exponentially on the positive axis, hence they do not satisfy the first bound. This follows from Stirling's approximation for $\Gamma(s)$ and the fact that $\zeta(\sigma+it)$ tends to $1$ for $\sigma\to\infty$ and $|t|<10$, say.

The functions $\xi(s)$ and $$f(s):=\xi\left(\frac{s-i}{2}\right) + \xi\left(\frac{i+s}{2}\right)$$ grow faster than exponentially on the positive axis, hence they do not satisfy the first bound. This follows from Stirling's approximation for $\Gamma(s)$ and the fact that $\zeta(s)$ tends to $1$ as $\Re s\to\infty$.

The entire functions satisfying the first display have order zero (by definition). As $\xi(s)$ has order one (which follows easily from Stirling's approximation for $\Gamma(s)$ and the convexity bound for $\zeta(s)$), it does not satisfy the first display.

The notation $f(\xi)$ is confusing, because $\xi$ is not a variable but a function. A better notation would be $$f(s):=\xi\left(\frac{s-i}{2}\right) + \xi\left(\frac{i+s}{2}\right).$$ The order of this entire function is also one, hence it does not satisfy the first display. Indeed, the order of $f(s)$ is at most one, because the order of $\xi(s)$ is at most one. On the other hand, the order of $f(s)$ is at least one, because it grows faster than exponentially on the positive axis (again by Stirling's approximation).

The functions $\xi(s)$ and $$f(s):=\xi\left(\frac{s-i}{2}\right) + \xi\left(\frac{i+s}{2}\right)$$ grow faster than exponentially on the positive axis, hence they do not satisfy the first bound. This follows from Stirling's approximation for $\Gamma(s)$ and the fact that $\zeta(\sigma+it)$ tends to $1$ for $\sigma\to\infty$ and $|t|<10$, say.