QUESTION
Prove that there exists a $T:V\rightarrow W$ such that $N(T)=V'\subset V$ and $R(T)=W'\subset W$
ATTEMPTED ANSWER
Let $V$ and $W$ be finite-dimensional vector spaces over $F$. Let $A=\{a_1,\dots,a_l\}$ be a basis for $V'\subset V$, let $B=\{a_1,\dots,a_l,b_1,\dots,b_m\}$ be a basis for $V$, let $C=\{c_1,\dots,c_n\}$ be a basis for $W'\subset W$, and let $D=\{c_1,\dots,c_n,d_1,\dots,d_p\}$ be a basis for $W$. Thus $N(T)=V'$ and $R(T)=W'$ means that
\begin{eqnarray} T(a_1)&=&0+\cdots +0\\ T(a_2)&=&0+\cdots +0\\ \vdots\\ T(a_l)&=&0+\cdots +0\\ T(b_1)&=&k_{11}c_1+k_{21}c_2+\cdots+k_{n1}c_n\\ T(b_2)&=&k_{12}c_1+k_{22}c_2+\cdots+k_{n2}c_n\\ \vdots\\ T(b_n)&=&k_{n1}c_1+k_{n1}c_2+\cdots+k_{nm}c_n, \end{eqnarray}
or simply
\begin{eqnarray} \begin{pmatrix} c_1&c_2&\cdots&c_n \end{pmatrix} \begin{pmatrix} 0&\cdots&0&k_{11}&\cdots&k_{n1}\\ \vdots&&\vdots&\vdots&&\vdots\\ 0&\cdots&0&k_{n1}&\cdots&k_{nm} \end{pmatrix}, \end{eqnarray}
which is an $(l+n)\times (l+m)$ matrix. Thus, such a $T$ exists and has the above form.
JUST CURIOUS
As far as I understand, we're dealing with something that looks like this:
I do not see the relevance of your attempted proof to the question you pose.
I'm guessing that $N(T)$ is the kernel of $T$ and $R(T)$ is the image of $T$. Let's look at what these sets are. By definition, we have
$$N(T) = \{v \in V : T(v) = 0\}$$
and
$$R(T) = \{ T(v) : v \in V\}$$
From these definitions, can you convince yourself that $N(T) \subseteq V$ and $R(T) \subseteq W$?