Let $Φ$ be a root system of type $G_2$, with base ${α, β}$ where $α$ is short. Then $Ψ_1 :=$ ±{$α, 3α + 2β$} is a closed subsystem of type $A_1A_1$, and clearly $|\mathbb{Z}Φ/\mathbb{Z}Ψ_1| = 2$; furthermore, $Ψ_2 = ±${$β, 3α + β, 3α + 2β$} is a closed subsystem of type $A_2$ with $|\mathbb{Z}Φ/\mathbb{Z}Ψ_2| = 3$, so $2$ and $3$ are certainly bad primes for $G_2$.
I can't get this: $|\mathbb{Z}Φ/\mathbb{Z}Ψ_2| = 3$. It's supposed to be the number in $\mathbb{Z}Φ$ which cannot be expressed as a linear combination of $\mathbb{Z}Ψ_2$, if I am correct? I only see $α,α + β$ .
Hint 1: Via the isomorphism $$\mathbb Z \Phi \simeq \mathbb Z \oplus \mathbb Z$$ $$n \alpha + m \beta \mapsto (n,m)$$
you are looking at the quotients of $\mathbb Z \oplus \mathbb Z$ with respect to the subgroups generated by $\{(1,0),(3,2)\}$ (for $\Psi_1$) and by $\{(0,1), (3,1),(3,2)\}$ (for $\Psi_2$), respectively. (Note also that the last subroup is already generated by $(0,1)$ and $(3,1)$ alone.)
Hint 2: The two elements of $\mathbb Z \Phi/ \mathbb Z\Psi_1$ are represented by $0$ and $\beta$. In $\mathbb Z \Phi/ \mathbb Z\Psi_1$, we have $2\beta \equiv -3\alpha \equiv 0$.
Hint 3: In $\mathbb Z \Phi/ \mathbb Z\Psi_2$ we actually have $\alpha \equiv \alpha+ \beta$ because $\beta \equiv 0$, so the two representatives you write in the question actually give the same element. Rather, since $\beta \equiv 0$, look at multiples of $\alpha$ alone and check how many there are up to equivalence in that quotient.