Let $f : G \rightarrow G'$ - surjective homomorphism, $H \vartriangleleft G$.
Prove that, $f(H) \vartriangleleft G'$.
My proof:
We have $H = aHa^{-1} \rightarrow f(H) = f(aHa^{-1})$. Let's denote $f(H)$ by $H'$, then $f(H) = H' = \{h' \in H' | f(h)\ = h'\}$.
$f(aHa^{-1}) = aH'a^{-1} = \{h' \in H' | f(h)\ = h'\}$.
Since, $f(H) = f(aHa^{-1}) \rightarrow H' = aH'a^{-1}$.
And in general case, what steps i should follow to show that some group is a normal subgroup of any other group?
You have a problem in your proof: $\;a\in G\;$ but it may well be that $\;a\;$ is not even an element of $\;G'\;$ ...!
But it doesn't matter: let $\;x\in G'$ be any element, then $\;x=f(g)\;$ , for some $\;g\in G\;$ (why?) , so that
$$xH'x^{-1}=f(g)f(H)f(g)^{-1}=f(gHg^{-1})=f(H)=H'$$
and you got $\;xH'x^{-1}=H'\;$ for any $\;x\in G'\iff H'\lhd G'\;$ and we're done.