As you've shown $Ker(\phi)=\{1,6\}$
You'd have $G/Ker(\phi)=\{\{1,6\},\{a,b\},\{c,d\}\}$
Note that $Ker(\phi)$ partitions the set $G$ in such a way that the corresponding pairs which are formed map to the same element under $\phi$.
Why?
By definition of cosets $G/Ker(\phi)=\{g.Ker(\phi)\,\,\big| \,\,g\in G\}$, then $\{a,b\}=\{g.1,g.6\}$ for some $g$.
Now $\phi(a)=\phi(g.1)=\phi(g)\phi(1)=\phi(g)\phi(6)=\phi(g.6)=\phi(b)$
Recall: $\phi(1)=1=\phi(6)$. Also note that $\phi(2)=4=\phi(5)$ and $\phi(3)=2=\phi(4)$ (NOTE that the operations still take place modulo $7$).
How did we reach here?
Well it is a well known fact that when we talk about ordinary addition $g^2=(-g)^2$
Now, same is the case here (with a little twist), what is the additive (modulo 7) inverse of $2$? $5$ is it? Of course, in fact additive inverse of $a$ (modulo 7) is $7-a$ and you have the same concept as the one used above.
$$a^2\equiv (7-a)^2\pmod7$$
So $\{a,7-a\}\in G/Ker(\phi)$ for $a\in H$ and the set becomes
$G/Ker(\phi)=\{\{1,6\},\{2,5\},\{3,4\}\}$
There is a more rigorous way that has been shown in most of the answers above, this is slightly different (intuitive?) way
Now, for the part where you ask what is $G/Ker(\phi)$ isomorphic to
Note $|G/Ker(\phi)|=|G|/|Ker(\phi)|=3$ and as it is a well known fact that any group (of quotients here) of prime order is cylic. So, $G/Ker(\phi)$ is isomorphic to any cyclic group of order 3. There is a theorem (Fundamental Theorem of Isomorphism) which states that
If $\phi:G\to G'$ (here G=G') is a homomorphism (Note that $G$ is onto $\phi(G)$), then $G/Ker(\phi)\cong\phi(G)$.
Now note, here $\phi(G)=\{1,2,4\}=\langle 4\rangle=\langle 2\rangle$ (modulo 7)
Best Answer
There is really "only one" group with three elements, $\mathbb Z/3\mathbb Z$. You just need to find out which element in your group is the neutral element and how the other two behave when multiplied, and you can easily write down a 1-1 correspondence, which if you verify that it is a group homomorphism will be your isomorphism.
Edit. Some very explicit details.
Let the three elements of the group $G/H$ be denoted $a,b,c$.
1. Investigating the structure.
Note that mod 7 we have $$\text{a}^2:\quad 1^2 = 6^2= 1$$ $$\text{b}^2:\quad3^2 = 4^2 = 2$$ $$\text{c}^2:\quad2^2 = 5^2 = 4$$ with the additional properties that $1*1=1$ and $2*2=4$ and $2*4= 1$.
Let's write these as $a*a=a$ and $b*b=c$ and $b*c=a$.
This is just like $\mathbb Z/3$ (with addition) where $0+0=0$ and $1+1=2$ and $1+2=0$.
2. Defining a bijection.
Let $f:G/H\to \mathbb Z/3$ be defined by $1\mapsto 0$, $3\mapsto 1$ and $2\mapsto 4$. It is obviously bijective. You just need to verify that it is a homomorphism!