Let $\phi: M\rightarrow N$ be a homomorphism. $M,N$ are left $R-\text {modules}$.
Prove that $\ker\phi<M$.
Fix $m_1,m_2\in \ker\phi\\\phi(m_1)=\phi(m_2)=0\\\phi(m_1+m_2)=\phi(m_1)+\phi(m_2)=0+0=0.$
So $m_1+m_2\in \ker\phi.$
Fix $a\in R\\\phi(am_1)=a\phi(m_1)=a\cdot0=0.$
So $am_1\in \ker\phi.$
Okay, I understand the transformations but why this implies that $\ker\phi$ is a submodule? What definition of submodule is used here?
Well, an $R$-submodule of an $R$-module $M$ is given by a nonempty subset $U\subseteq M$ which forms with the restrictions of the operations on $M$ (addition and scalar multiplication) to $U$ itself an $R$-module.
This definition can be used to define the substructure of any algebraic structure.
What you have proved is that $U=\ker\phi$ is closed under addition and scalar multiplication. This is sufficient since the restriction of addition and scalar multiplication to $U$ fulfull all axioms/rules for being an $R$-module.