The addition in a quotient group $\mathbb{Z}_{15}/\langle 6\rangle$ is defined as
$$(a+\langle 6\rangle) + (b + \langle 6\rangle) = (a+b) + \langle 6 \rangle$$
for $a,b \in \mathbb{Z}_{15}$. Hence no adding of $\langle 6\rangle$s.
On the other hand, equality in the quotient group is defined as $a + \langle 6\rangle = b + \langle 6\rangle$ if $a-b \in \langle 6\rangle$. The identity element of $\mathbb{Z}_{15}/\langle 6\rangle$ is $0 + \langle 6\rangle$, which is commonly denoted simply as $\langle 6\rangle$.
Let's determine $\langle 6\rangle$. Certainly $0,6 \in \langle 6\rangle$. Since a subgroup is closed under inverting we also have $-6 \in \langle 6\rangle$. But we have $-6 = 9$ in $\mathbb{Z}_{15}$ (the equality actually being congruence mod $15$) so $9 \in \langle 6\rangle$.
Since a subgroup is closed under addition, we have $12 = 6 + 6 \in \langle 6\rangle$. And then again $3 = -12 \in \langle 6\rangle$.
So far we got $\{0,3,6,9,12\} \subseteq \langle 6\rangle$. Now check that $\{0,3,6,9,12\}$ is indeed a subgroup of $\mathbb{Z}_{15}$, i.e. that the inverse of every element and the product of every two elements from the set stays in the set (working mod $15$ of course). Since $\langle 6\rangle$ is the smallest subgroup of $\mathbb{Z}_{15}$ containing $6$, we conclude $\langle 6\rangle \subseteq \{0,3,6,9,12\}$, and hence they are equal.
Finally, we see that $3 + \langle 6\rangle = 0 + \langle 6\rangle$ because $3-0 = 3 \in \langle 6\rangle$. Therefore $3 + \langle 6\rangle$ is indeed the identity element $\langle 6\rangle$ in $\mathbb{Z}_{15}/\langle 6\rangle$.
Best Answer
You're wrong on what $B$ is. It is the set generated by both $2$ and $6$ individually. So instead
$$B = \left\{ 2n + 6m \mid n,m \in \Bbb Z \right\}$$
Of course, since a multiple of $6$ is a multiple of $2$ as well, we see $A=B$.