Intuitive Explanation of Morphism Theorem Is there an intuitive explanation for the morphism theorem from introductory abstract algebra?

First Morphism Theorem: Let $K$ be the kernel of the group morphism $f: G \to H$.  Then $G/K$ is isomorphic to $\text{Im } f$, and the isomorphism $\psi : G/K \to \text{Im } f$ is defined by $\psi(Kg) = f(g)$.

I follow the steps of the proof of the theorem, but find the theorem very nonintuitive to conceptualize. 

Quotients bunch up elements that are the same in some way, up into one thing. Quotienting $\mathbb{Z}$ by the subgroup $3\mathbb{Z}$ puts all things that have the same remainder after dividing by 3, into one set. So -5, 1 and 7 are all different elements in $\mathbb{Z}$, and the coset $$\{-5,-2,1,4,7\cdots\}=1+3\mathbb{Z}$$ puts everything  with remainder 1 after division by 3, into one set.
Now let's try to make a morphism $f:G\to H$ into an isomorphism. We can make it surjective simply by restricting the image. So $f^*:G\to \text{Im }f$ given by $f^*(g)=f(g)$ is a surjective homomorphism.
But how do we make it injective? Many elements are getting mapped to the same image, because for any element $k$ of the kernel, $f(kg) = f(g).$ In fact, the pre-image of $f(g)$ is $Kg.$ To make it injective, we could put everything that gets mapped to $f(g)$ together so that by construction, only one thing gets mapped to each element of the image (i.e. being injective). This means precisely considering the quotient $G/K.$ So now if we have $\phi:G/K \to \text{Im } f$ defined by $ \phi(Kg) = f(g)$ then it is certainly surjective, a morphism and injective as well because we clumped all the things that used to get sent to $f(g)$ all up into one element.
