How can I use the master theorem to get an upper bound on this recurrence? I Wrote an algorithm whose running time is described by the recurrence
$$T(n) = 2T(n/2) + \Theta(n\lg n),$$
and I want to determine whether $T(n) = o(n^2)$.
Now by the third case of the "Master Theorem" a recurrence of the form
$$T(n) = 2T(n/2) + f(n)$$
has solution $T(n) = \Theta(f(n))$, if $f(n) = \Omega(n^{1+\varepsilon})$ for some $\varepsilon > 0$.
I can't apply this directly to my recurrence since $n\lg n \neq \Omega(n^{1+\varepsilon})$, but if I can find a function $f(n)$ such that
$$n\lg n = \Omega(f(n))\qquad \text{ and } \qquad f(n) = \Omega(n^{1+\varepsilon}) \qquad\text{ and } \qquad f(n) = o(n^2),$$
then I can just use the master theorem on
$$T(n) = 2T(n/2) + f(n)$$
to get the bound. So is there such an $f(n)$?

None of the Master Theorem's cases apply to your problem. In order to find the asymptotic running time of your recurrence, you need an extended version described in exercise 4.6-2 of CLRS book (extension of Case 2). Besides the Wikipedia to which Ian already referred, see also here or here for a proof of this case extension.
So you have $a=b=2, k=1$, and you get $T(n)=\Theta(n\lg^2n)=o(n^2)$
