The $SU(2)$ triplet results from the Adjoint Representation $\mathrm{Ad}: SU(2)\to SO(3)$ of $SU(2)$, whereby $SU(2)$ acts on its own Lie algebra. As a $2\times2$ matrix, an element of the Lie algebra $\mathfrak{su}(2)$ can be written:
$$X=\left(\begin{array}{cc}i\,z&i\,x - y\\i\,x + y&-i\,z\end{array}\right)=i\,(x\,\sigma_x+y\,\sigma_y + z\,\sigma_z)\tag{1}$$$$X=\begin{pmatrix}i\,z&i\,x + y\\i\,x - y&-i\,z\end{pmatrix}=i\,(x\,\sigma_x+y\,\sigma_y + z\,\sigma_z)\tag{1}$$
where $\sigma_j$ are the Pauli matrices and $\gamma\in SU(2)$ acts on this entity through the so-called spinor map $X\mapsto\gamma\,X\gamma^{-1}$. So the triplet is a 3 element, real vector of the co-efficients of the Pauli matrices in (1) and, to find the matrix of the triplet transformation, you need to find the matrix of the linear, homogeneous transformation $X\mapsto\gamma\,X\gamma^{-1}$.
If you have the $SU(2)$ member in the form you write, i.e. $\gamma =\exp\left(i\frac{\sigma_i}{2}\theta_i\right)$ then there is an easy way to find the matrix of the triplet transformation since we can show that (look up the relationship between the adjoint representation of the Lie algebra and that of the group) $X\mapsto \exp\left(i\frac{\sigma_i}{2}\theta_i\right)\, X\, \exp\left(-i\frac{\sigma_i}{2}\theta_i\right)$ is the transformation:
$$\left(\begin{array}{c}x\\y\\z\end{array}\right)\mapsto\exp\left(\frac{1}{2}\theta_i \,\mathrm{ad}(i\,\sigma_i)\right)\left(\begin{array}{c}x\\y\\z\end{array}\right)\tag{2}$$
where $\mathrm{ad}(Y)$ is the $3\times3$ matrix of the linear mapping $X\mapsto[Y,\,X]$. So you need to work out the $\mathrm{ad}(i\,\sigma_i)$ from the commutation relationships for the Pauli matrices. The result is the triplet transformation law:
$$\left(\begin{array}{c}x\\y\\z\end{array}\right)\mapsto\exp\left(\left(\begin{array}{ccc}0&-\theta_z&\theta_y\\\theta_z&0&-\theta_x\\-\theta_y&\theta_x&0\end{array}\right)\right)\left(\begin{array}{c}x\\y\\z\end{array}\right)\tag{3}$$
