Let $J$ is ideal of $R={\Bbb Z_{5}[x]/}{\langle f(x)\rangle }$
Prove that there exists $g(x)$ such that $J={\langle g(x)\rangle/}{\langle f(x)\rangle}$ and $\langle f(x)\rangle \subseteq \langle g(x)\rangle $
Proof:
Let $\Phi$ be the map on $\Bbb{Z}_{5}[x]$ defined by $\Phi(g(x))=g(x)+\langle f(x)\rangle$, then $\Phi$ is onto and ring homomorphism.
Suppose $J$ is ideal of $R$, then $\Phi^{-1}(J)$ is ideal of $\Bbb Z_{5}[x]$ so there exists $h(x)$ such that $\Phi^{-1}(J)=\langle h(x)\rangle$.
Since $\Phi$ is onto, $J=\Phi(\Phi^{-1}(J))=\Phi(\langle h(x)\rangle)$
My question:
How to show $\Phi(\langle h(x)\rangle)={\langle h(x)\rangle/}{\langle f(x)\rangle}$ and $\langle f(x)\rangle\subseteq\langle h(x)\rangle$ ?
To answer your question, to see that $(f)\subseteq (h)$, note that $f\in\Phi^{-1}(J)$, since $\Phi(f)=0\in J$, so $f\in (h)$, which implies that $(f)\subseteq (h)$.
As for your other question, since $\Phi:\Bbb{F}_5[x]\to \Bbb{F}_5[x]/(f)$ is the quotient map, $\Phi((h))=(h)/(f)$, pretty much by definition of $(h)/(f)$. You'd have to be a bit more clear about what's confusing you here for me to know how to explain this further.