Quotient Ring with kernel?

The way this theorem is interpreted is that, for any ideal $I$ of the ring $R$, you can form a homomorphism from $R$ to $R/I$ by sending an element $x\in R$ to the coset $x+I$ in $R/I$. The kernel of the homomorphism is $I$.

Neither of your remarks are really related to the theorem, although they do have some meaning. First of all, notice that a quotient ring $R/I$ is defined by $I$ being an ideal. So for any ideal $I$ of $R$, you can form a quotient ring, and that construction/definition has nothing to do with this theorem. (You would need the definition of $R/I$ before stating this theorem). A quotient ring a priori has nothing to do with a homomorphism or a kernel, so your statement "$R/I$ exists iff $I$ is kernel" has no meaning.

Another thing to note is that the kernel of a homomorphism is always an ideal, which is a good exercise. However, you can have a homomorphism $R\to R/I$ that doesn't have a kernel of $I$.