The homomorphism proof as it currently stands is incorrect.
You have:
\begin{align*}
\phi(n_1)+\phi(n_2) & \color{red}{=} \phi(n_1+n_2) \\
& \color{red}{=} \phi(n_1)+\phi(n_2) \\
& = 7n_1+7n_2 \\
& = 7(n_1+n_2)
\end{align*}
You have the important elements here, but they're arranged in a confusing way, that I would consider incorrect.
Specifically, the red equalities above are incorrect.
If you have a group $(G,+)$, and another group $(H,\oplus)$, a group homomorphism $\phi:G\to H$ is a function $\phi$ such that, for all $x,y\in G$:
$$\phi(x+y) = \phi(x)\oplus\phi(y)$$
Note that there are different operations "inside" $\phi$'s brackets vs outside of them.
This is because the elements $x,y\in G$ (so are combined using $G$'s operation), but $\phi(x),\phi(y)\in H$ (so are combined using $H$'s operation).
A homomorphism is a function that satisfies the above.
So, to prove something is a homomorphism, you need to prove that it always satisfies that equation.
Let $\phi:\mathbb Z\to\mathbb Z$ be defined by $x\mapsto 7x$.
Then, we have that $\phi(x+y) = 7(x+y)$.
We also have that $\phi(x)+\phi(y) = 7x+7y$.
Can we show that these expressions are equal? If we can, then $\phi$ is a homomorphism.
Showing they're equal isn't too bad, what you do is say that:
$$\phi(x+y) = 7(x+y) \color{red}{=} 7x+7y = \phi(x)+\phi(y)$$
Everything here I've written is just "writing what $\phi(x)$ means" besides the equality in red, which using distributivity of addition in $\mathbb Z$.
But, what we've done here is start with $\phi(x+y)$, and show how that must be equal to $\phi(x)+\phi(y)$, so $\phi$ must be a homomorphism.
Best Answer
The order of $G$ is 24 [why is this], so plugging this into your equation $|G| =$ $|$ker$(f)| \times |$Im$(f)|$ yields $24 =$ $|$ker$(f)| \times 2$ or $|$ker$(f)| = 12.$