The trouble is that you don't *know* $f$; it could be any one of the functions in $\Theta(n^2)$ resp. $\Omega(n^2)$.

So let's unfold the classes! In case one, whichever function $f \in \Theta(n^2)$ happens to be the "real" one, we know that there is a $d_1 \in \mathbb{N}$ such that 

$\qquad \displaystyle f(n) \leq d_1n^2$,

and also a $d_2 \in \mathbb{N}$ such that

$\qquad \displaystyle f(n) \geq d_2n^2$

for all $n \in \mathbb{N}$ (that are the $O$- resp. $\Omega$-part of $\Theta$). We have to show that there is some constant $c$ such that

$\qquad \displaystyle 2f(n/2) \leq cf(n)$.

With above inequalities, we get that $2f(n/2) \leq \frac{1}{2}d_1n^2$, and that $cd_2n^2 \leq cf(n)$. So, if we choose $c$ such that

$\qquad \displaystyle \frac{1}{2}d_1n^2 \leq cd_2n^2$,

we have shown the necessary inequality. Note that $c=d_1$ already works ($d_1,d_2 > 0$).

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As you see, we needed *both* upper and lower bound to perform the proof. We don't have that in the second case where $f \in \Omega(n^2)$. That suggests the statement may be false (but does not prove it).

Once we suspect, we can just choose a counter-example. However, I have not been able to find one! Instead, some handwaving thinking suggests the following:

> **Conjecture**  
> For $f\in \Omega(n^2)$, $f\left(\frac{n}{2}\right) < \frac{f(n)}{2}$.

This is quite plausible; I am not sure what wicked functions may do to the fraction, though. If this is the case, the factor $2$ is not able to compensate for the increase in the divisor $2$, and $f(n)$ dominates whole $T(n)$.

This line of reasoning is wrong, though. (It *may* be correct for "smooth" functions). See [Yuval's answer](http://cs.stackexchange.com/a/4980) for a counter-example.