Why if $f:A\rightarrow A_{1}$ is an $R$-module homomorphism then $f(B) \subset B_{1} $ and $f(C) \subset C_{1}.$? The question is:

If $R$ has an identity and $A$ is an $R$-module, then there are submodules $B$ and $C$ of $A$ such thatt $B$ is unitary $RC = 0$ and $A = B \oplus C.$
And if $A_{1}$is another $R$-module, with $A_{1} = B_{1} \oplus C_{1}$, $B_{1}$ unitary ,$RC_{1} = 0.$

Why if $f:A\rightarrow A_{1}$ is an $R$-module homomorphism then $f(B) \subset B_{1} $ and $f(C) \subset C_{1}.$?
I assume your modules are not necessarily unital. Let $M$ be a left $R$-module; define
$$
z(M)=\{x\in M:1x=0\}\qquad u(M)=\{x\in M: 1x=x\}
$$
and prove $z(M)$ and $u(M)$ are submodules of $M$, with $Rz(M)=0$.

Now let $x\in M$ and consider $x=(x-1x)+1x$; then
$$
1(x-1x)=1x-1x=0
\qquad
1(1x)=1x
$$
so $x-1x\in z(M)$ and $1x\in u(M)$, so $M=z(M)+u(M)$. Also, $x\in z(M)\cap u(M)$ implies $1x=0$ and $x=1x$, so the sum is direct.

Now prove that for a homomorphism $f\colon M\to N$ you have $f(z(M))\subseteq z(N)$ and $f(u(M))\subseteq u(N)$.
Hint:Assume that $f(B) \not\subset B_{1}$. We have some $0\neq b\in B$ such that that $f(b)\in C_{1}$ (i.e., $f(b)\notin B_1$). So, $f(b)=f(1_Rb)=1_Rf(b)\in C_1$.But, we know that $RC_1=0$, so $1_Rf(b)=f(b)=0$. This is a contradiction of $b\neq 0$. ($f(b)=0=f(0)$).
You can use above idea for $f(C) \subset C_{1}$.