Indeed, a triplet $\vec{\phi}$ transforming under the triplet representation matrices T can be doted to a Pauli vector to yield a formal adjoint, so a traceless 2×2 matrix $$\Phi=\tfrac{1}{2}\vec{\phi}\cdot \vec{\tau}~,$$ transforming as the doublet representation, conjugately on both sides. This is dubbed adjoint action,
$$
\Phi \mapsto e^{i\vec{\tau}\cdot \frac{\vec{\theta}}{2}}~ \Phi ~e^{-i\vec{\tau}\cdot \frac{\vec{\theta}}{2}}.
$$
In your case,
\begin{equation}
\Delta=\begin{pmatrix} \frac{\Delta^{+}}{\sqrt{2}} & \Delta^{++} \\
\Delta^0 & - \frac{\Delta^{+}}{\sqrt{2}}\end{pmatrix}= \sqrt{2}\Delta^+ \frac{\tau_3}{2} +\Delta^{++} \frac{\tau_1+i\tau_2}{2} + \Delta^0 \frac{\tau_1-i\tau_2}{2} = \sqrt{2}\Delta^+ \frac{\tau_3}{2} +\Delta^{++} \frac{\tau_ +}{2} + \Delta^0 \frac{\tau_-}{2} \\ = \begin{pmatrix}\Delta^{++}+\Delta^0\\i(\Delta^{++}-\Delta^0)\\\sqrt{2}\Delta^+ \end{pmatrix}\cdot \frac{\vec{\tau}}{2},
\end{equation}
where you note the normalization
$$
\vec{\bar{\phi}}\cdot \vec{ \phi }=2(\Delta^{++~~2}+\Delta^{+~~2}+\Delta^{0~~2}).
$$
You can easily see, then, that the Cartesian 3-vector can be unitarily rotated to the more transparent spherical basis vector $\sqrt{2}(\Delta^{++},\Delta^+, \Delta^0)$.
The matrix $U^\dagger$ achieving that Cartesian to spherical basis change is given in footnote nb 3 of the WP article
or this answer . In an innocuously rephased form,
$$
U^\dagger \begin{pmatrix}\Delta^{++}+\Delta^0\\i(\Delta^{++}-\Delta^0)\\\sqrt{2}\Delta^+ \end{pmatrix} = \frac{1}{\sqrt{2}} \left(\begin{matrix} 1 & -i & 0 \\ 0 & 0 & \sqrt{2} \\ 1 & i & 0\end{matrix}\right) ~\begin{pmatrix}\Delta^{++}+\Delta^0\\i(\Delta^{++}-\Delta^0)\\\sqrt{2}\Delta^+ \end{pmatrix}= \sqrt{2}\begin{pmatrix}\Delta^{++}\\\Delta^+ \\ \Delta^0 \end{pmatrix}.
$$
To confirm the normalization of the half-angle and the signs, take only the 3rd component of θ to be non-vanishing and infinitesimal, so the rotation increment of Δ is a simpler matrix with vanishing diagonals. Check that the increment components of $\vec{\phi}$ are now given by commutators (adjoint) and fully comport with the classical cross-product increment of the triplet representation you specified. It is then evident upon commutation with $\tau_3/2$ that the $T_3$ eigenvalues of the triplet $(\Delta^{++},\Delta^+, \Delta^0)$ are (1,0,-1), so that its $Y=Q-T_3=1$. I normalize the weak hypercharge the modern ("alternative") way, i.e. dropping the superfluous strong denominator of 2.
You may find the derivation of T to all orders through this construction in this answer.
