An anti-particle transforms in the conjugate transformation $\bar{\mathbf r}$ than the particle ${\mathbf r}$ does under the given symmetry group, $SU(2)$ in your case. No surprise thus in general that the generators $T^i_{\mathbf r}$ and $T^i_\bar{\mathbf r}=-T^{i\,*}_{\mathbf r}$ act differently on ${\mathbf r}$ and $\bar{\mathbf r}$ respectively.
This is the reason why for e.g. spin-$\frac{1}{2}$ fields, that is linear combinations of particle annihilation operators, $a$, and anti-particles creation operators, $b^\dagger$, the wave functions $u(p)$ and $v(p)$ in front to $a$ and $b^\dagger$ are not symply related by $p\rightarrow -p$.
However, for SU(2) doublet transforming with generators $T^i=\sigma^i/2$ (where $\sigma^i$ are the Pauli matrices), it turns out that $\sigma^2 T^i\sigma^2=-T^{i\,*}$ so that $i\sigma^2 \bar{\mathbf{2}}$ transform as a ${\mathbf 2}$, namely
$$
i\sigma^2\psi^*=\left(\begin{array}{c} \bar{d}\\ -\bar{u}\end{array}\right)\qquad \mbox{and}\qquad \psi=\left(\begin{array}{c} u\\ d\end{array}\right)
$$
transform in the same way under $SU(2)$ (and in particular $\tau_+ \bar{u}=-\bar{d}$).
Therefore, if you want to extract a triplet from $\bar{\mathbf 2}\otimes {\mathbf 2}={\mathbf{1}}+\mathbf{3}$, you just need to take the symmetric combinations of $i\sigma^2_{ab} \psi^{b\,*} \psi^c + i\sigma^2_{cb} \psi^{b\,*} \psi^a={\mathbf{3}}$, exactly like if you wanted to extract the triplet (i.e. the symmetric combination) out of $\mathbf 2\otimes {\mathbf 2}={\mathbf{1}}+\mathbf{3}$.
Explicitly, ordering the states as $\psi^{a=1}=|up\rangle=u$ and $\psi^{a=2}=|down \rangle=d$ as above (this is the basis where $T^3$ is diagonal with $+1/2$ and $-1/2$ on the upper and lower entry respectively), the symmetric combination with $a=1$ and $c=2$ in $i\sigma^2_{ab} \psi^{b\,*} \psi^c + i\sigma^2_{cb} \psi^{b\,*} \psi^a$ is proportional to $\bar{u}u-\bar{d}d$. The relative minus sign that you get comes from the $i\sigma^2$. The other states of the triplet correspond to the choice $a=c=1$ and $a=c=2$, that give $\bar{d}u$ and $\bar{u}d$
$$
\mathbf{3}\propto \frac{1}{2}\left(i\sigma^2_{ab} \psi^{b\,*} \psi^c + i\sigma^2_{cb} \psi^{b\,*} \psi^a\right)=\left(\begin{array}{cc}\bar{d}u & \frac{1}{2}(\bar{d}d-\bar{u}u) \\
\frac{1}{2}(\bar{d}d-\bar{u}u) & -\bar{u}d
\end{array}\right)_{ac}
$$
This are the correct composition indeed, see http://en.wikipedia.org/wiki/Pion to check it out.
If you remove, erroneously, the $i\sigma$ form the symmetric combination, you get instead $\bar{u}d+\bar{d}u$ (for $a=1$, $c=2$), and $\bar{u}u$ and $\bar{d}d$ for the other choices $a=c$, a result that does not make any sense.
