The correspondence theorem for rings states:
Let A be a multiplicative ring with identity and I an ideal of A. There is a one-to-one correspondence between the ideals of A that contain I and the ideals of the quotient ring A/I.
In my lecture notes, it states:
If $\varphi\ :\ R \rightarrow S$ is an homomorphism and $I \trianglelefteq R$, then $\varphi(I)$ isn't necessarily an ideal of $S$. However, if $\varphi$ is surjective and $\ker(\varphi) \leq I$ then the correspondence theorem implies $\varphi(I) \trianglelefteq S$.
It's unclear to me how does one imply the other. If $\varphi$ is surjective, then $\forall s \in S\ \exists r \in R\ :\ s=\varphi(r)$ hence $\forall s \in S \quad s \varphi(I)=\varphi(rI)=\varphi(I)$, thus $\varphi(I) \trianglelefteq S$. How is the correspondence theorem relevant here?
You proved the assertion that if $\varphi$ is surjective then $\varphi(I)$ is an ideal, directly, by definition.
What the lecture notes say is it can be proved, indirectly, by the (strong form of) correspondence theorem and the isomorphism theorem. Namely if $\varphi$ is surjective then there is the induced ring isomorphism $\bar\varphi \colon R/\ker\varphi \to S.$ Thus the correspondence theorem implies that an ideal $I$ that contains $\ker\varphi$ corresponds to the ideal $\pi(I)$ of $R/\ker\varphi$ where $\pi \colon R \to R/\ker\varphi$ is the canonical map. Since $\varphi(I) = \bar\varphi(\pi(I))$, we have $\varphi(I)$ is an ideal as well.
Note that in the second proof, we need strong form of the correspondence theorem than the form you've cited: Let $\pi \colon A \to A/I$ be a canonical map. Then there is the ono-to-one correspondence between ideals of $A$ that contains $I$ and ideals of $A/I$ defined by $J \mapsto \pi(J)$.