I try to prove the following theorem which given without proof here On Prime Ideals of Lie Algebras
Theorem: Let $L$ and $L^{\prime}$ be Lie algebras and let $f: L \rightarrow L^{\prime}$ be a surjective homomorphism. Then an ideal $P$ of $L$ containing $\mathrm{Ker} f$ is prime if and only if $f(P)$ is prime in $L^{\prime}$.
My Proof: Let $P$ be a prime ideal of $L$, let $H$ and $K$ are two ideals of $L^{\prime}$. Suppose that $[H,K] \subseteq f(P)$, then $f^{-1}([H,K]) \subseteq f^{-1}f(P)$, hence $\Big [f^{-1}(H),f^{-1}(K)\Big] \subseteq P$. But $P$ is prime, then $f^{-1}(H) \subseteq P$ or $f^{-1}(K) \subseteq P$. Since $f$ is surjective homomorphism, thus $H \subseteq f(P) \textit{ or } K \subseteq f(P)$. Therefore $f(P)$ is prime.
Is this proof true?
Is this step $f^{-1}([H,K])= \Big [f^{-1}(H),f^{-1}(K)\Big]$ true?
Is this step $f^{-1}f(P)=P$ true?
Is this theorem is a trivial conclusion of Isomorphism Theorems of Lie algebra and we do not need this proof?
You have correctly identified the weak points of your argument:
(1) Is this step $f^{-1}([H,K])= \Big [f^{-1}(H),f^{-1}(K)\Big]$ true?
No, in general this is not true. If the Lie bracket on $L$ was zero, then the right hand side would be $0$, but the left hand side would contain ker$(f)$.
However for your purpose all you need is: $$f^{-1}([H,K])\supseteq \Big [f^{-1}(H),f^{-1}(K)\Big],$$ which is true. To see this just apply $f$ to an element of the right hand side, and clearly you will land in $[H,K]$.
(2) Is this step $f^{-1}f(P)=P$ true?
Yes, but only because of one specific bit of information you were given, which you should mention to justify the statement. Namely ker$(f)\subseteq P$. Thus if $x\in f^{-1}f(P)$ then $f(x)=f(p)$ for some $p \in P$, and $$x=(x-p)+p,$$ with $p\in P$ and $x-p\in$ker$(f)\subseteq P$.