I have three questions :
I understand the first isomorphic theorem, which states that a homomorphic image of a group is isomorphic to the quotient group formed by the group $G$ and the kernel of group $G$.
$1$. Is this theorem only true for Kernel $K$, or for any normal subgroup of $G$ ?
Also, suppose there exists a homomorphism $\phi$ between $G$ and $G'$. Let $$H = \{x \in G \; ; \; \phi(x) \in H'\}.$$ Then $H$ is subgroup of $G$. We can also show that given that $H'$ is normal in $G'$, $H$ is normal in $G$. Here, there exists an homomorphism between $H$ and $H'$.
$2$. Is the function defining homomorphism between $G$ and $G'$ same as $H$ and $H'$ ?
From first isomorphism theorem, we can say, $G/K \cong G'$ and $H/K \cong H'$
$3$. Then Can I make this statement : Given a group $G$, and subgroup $H$ of $G$, if there exists a homomorphism between $G$ and $G'$ with Kernel $K$ and $H'$ being a subgroup of $G'$, such that $G/K \cong G'$ and $H/K \cong H'$, then $H$ is normal in $G$ and $H'$ is normal in $G'$ ?
To answer your first question, the kernel depends on the homomorphism in question, with a different homomorphism you'll have a different kernel. However, given any normal subgroup $H$, you can always find a homomorphism to some group whose kernel is $H$, namely, $\phi\colon G\to G/H$ is a surjective homomorphism with kernel $H$.
I don't quite understand your second question. I'm assuming you have a homomorphism $\phi\colon G\to G'$ and $H'\lhd G'$. If you are taking $H = \phi^{-1}(H')$, then $H$ is the kernel of $$G\xrightarrow{\phi} G'\to G'/H',$$ hence it is normal. You can restrict $\phi$ to $H$ to get a surjective homomorphism $\phi\colon H\to H'$ and from the first isomorphism theorem, $H/K\cong H'$.
Lastly, your claim is not true. Simply take a subgroup $H$ which is not normal, and the identity map $G\to G$. Then the kernel $K$ is trivial, $G/K\cong G, H/K\cong H$ but $H$ is not normal.