[...] $\Delta^+ \rightarrow p + \pi^0$, [...] $\Delta^+ \rightarrow n + \pi^+$,
which process is favored: the proton and neutral pion or neutron and charged pion [?]
Since the kinematics (and corresponding "phase space" factors) for the two final states are presumably as good as equal, the evaluation of the branching ratio
$$\text{BR} := \frac{\Gamma[ \Delta^+\rightarrow p+\pi^0 ]}{\Gamma[ \Delta^+\rightarrow n+\pi^+ ]}$$
simplifies to determining the ratio of "state constituent" transition probabilities
$$\text{BR} := \frac{\Gamma[ \Delta^+\rightarrow p+\pi^0 ]}{\Gamma[ \Delta^+\rightarrow n+\pi^+ ]} \simeq \frac{\left\lvert \langle p; \pi^0 \mid \Delta^+ \rangle \right\rvert^2}{\left\lvert \langle n; \pi^+ \mid \Delta^+ \rangle \right\rvert^2}.$$
Analyzing (or defining) the initial state $\Delta^+$ and the two distinct final states in terms of isospin leads to the expressions
$$ \lvert \Delta^+ \rangle \equiv \big\lvert \left(3/2, 1/2\right)_i \big\rangle, $$
where the first value represents the magnitude of $\mathbf I$, and the second value represents the magnitude of $I_3$, along with
$$ \lvert p; \pi^0 \rangle \equiv \big\lvert (1/2, 1/2)_f; (1, 0)_f \big\rangle \equiv \sqrt{ \frac{2}{3} }~\big\lvert (3/2, 1/2)_t \big\rangle - \sqrt{ \frac{1}{3} }~\big\lvert (1/2, 1/2)_t \big\rangle, $$ and
$$ \lvert n; \pi^+ \rangle \equiv \big\lvert (1/2, -1/2)_f; (1, 1)_f \big\rangle \equiv \sqrt{ \frac{1}{3} }~\big\lvert (3/2, 1/2)_t \big\rangle + \sqrt{ \frac{2}{3} }~\big\lvert (1/2, 1/2)_t \big\rangle, $$
where
the coefficients of the linear combinations on the right-hand sides are Clebsch-Gordan coefficients (specificly those values listed in table "$1/2 \otimes 1$"),
all states are normalized, and
the indices $f$ and $t$ are to distinguish final states and "state representations to evaluate transition probabilities"; such that in particular the states $(1/2, 1/2)_f$ and $(1/2, 1/2)_t$ are (meant to be) distinct; and both are distinct, and indeed disjoint, from the initial state $\lvert \Delta^+ \rangle \equiv \lvert (3/2, 1/2)_i \rangle$.
Now identifying
$$\big\lvert (3/2, 1/2)_t \big\rangle \equiv \big\lvert (3/2, 1/2)_i \big\rangle $$
we can evaluate
\begin{align}
\langle p; \pi^0 \mid \Delta^+ \rangle & \equiv \bigg\langle \sqrt{ \frac{2}{3} }~ (3/2, 1/2)_t - \sqrt{ \frac{1}{3} }~ (1/2, 1/2)_t \bigg\vert (3/2, 1/2)_t \bigg\rangle \\
& = \bigg\langle \sqrt{ \frac{2}{3} }~ (3/2, 1/2)_t \bigg\vert (3/2, 1/2)_t \bigg\rangle \\
& = \sqrt{ \frac{2}{3} }
\end{align}
and
\begin{align}
\langle n; \pi^+ \mid \Delta^+ \rangle & \equiv \bigg\langle \sqrt{ \frac{1}{3} }~ (3/2, 1/2)_t + \sqrt{ \frac{2}{3} }~ (1/2, 1/2)_t \bigg\vert (3/2, 1/2)_t \bigg\rangle \\
& = \bigg\langle \sqrt{ \frac{1}{3} }~ (3/2, 1/2)_t \bigg\vert (3/2, 1/2)_t \bigg\rangle \\
& = \sqrt{ \frac{1}{3} }
\end{align}
obtaining the sought branching ratio value as
$$\text{BR} := \frac{\Gamma[ \Delta^+\rightarrow p+\pi^0 ]}{\Gamma[ \Delta^+\rightarrow n+\pi^+ ]} \simeq \frac{ (\sqrt{ 2/3 })^2 }{ (\sqrt{ 1/3 })^2} = 2.$$