Computing images of $K$ in $G/N$ using the second isomorphism theorem. Second Isomorphism Theorem: Let $G$ be a group and $K,N \leq G$ with $N \unlhd G$. Then $KN/N \cong K/(K \cap N)$.
In lectures we mentioned that the second isomorphism theorem is useful for "computing" the image of $K$ in $G/N$ up to isomorphism.
But I don't really understand what that means. The image of $K$ under what? The natural projection homomorphism?
Example:    Let $G = \langle g \rangle$ and $|g| = 12$. Let $N = \langle g^3 \rangle  = \{1,g^3,g^6,g^9\}$ and $K = \langle g^2 \rangle  = \{1,g^2,g^4,g^6,g^8,g^{10}\}$. Then the image of $K$ in $G/N$ is isomorphic to $K/(K\cap N) = \langle g^2 \rangle/\langle g^6 \rangle$. But again image of $K$ under what, and how do we jump to the fact that it's isomorphic to $K/(K\cap N)$?
Thanks in advance for the clarification.

Image of $K$ under the canonical projection to $G/N.$ And $KN$ is the set of $g^{2 a + 3 b}$ for every integer $a, b$. Since $2$ and $3$ are relatively prime, this is all of $G$ (seen more simply by setting $a=-1, b=1$). So, $G/N = K/(K\cap N),$
