Is the following Proof Correct? In particular i am concerned with the proof of the forward implication.
Theorem. Given that $W$ is finite-dimensional and $T\in\mathcal{L}(V,W)$. then $T' = 0$ if and only if $T = 0$.
$(\Rightarrow).$ Assume that $T' = 0$ and $v\in V$. Let $w_1,w_2,...,w_m$ be a basis for $W$ and $\phi_1,\phi_2,...,\phi_m$ be the corresponding dual basis. Expressing $Tv$ as a linear combination of $w_1,w_2,..,w_m$ we have $$Tv = c_1w_1+c_2w_2+\cdot\cdot\cdot+c_mw_m$$ From hypothesis $T' = 0$ so equivalently given any linear functional $\psi\in W'$, $T'(\psi) = \psi\circ T = 0$, then in particular $\phi_1\circ T =\phi_2\circ T =\cdot\cdot\cdot= \phi_m\circ T = 0$. Now let $j\in\{1,2,...,m\}$ , taking together our immediate deduction with the equation above we have $$0= (\phi_j\circ T)v = \phi_j(Tv) = \phi_j(c_1w_1+c_2w_2+\cdot\cdot\cdot+c_mw_m) = c_j$$ Since our choice of $j$ was arbitrary it follows that $c_1 = c_2 = \cdot\cdot\cdot = c_m = 0$ and by implication $Tv = 0$ moreover since $v$ was arbitrary it follows that $T = 0$.
$(\Leftarrow).$ Assume now that $T = 0$ and let $\phi\in W'$ by definition $T'(\phi) = \phi\circ T$ since $T = 0$ it follows that $\forall v\in V(Tv = 0)$ consequently $\phi(Tv) = 0$ thus $T'(\phi)$ is the 0 linear functional in $V'$ implying that $T'=0$
$\blacksquare$
The proof is overall correct, but you can make it simpler.
Suppose $T\ne0$ and let $v\in V$ with $w_1=T(v)\ne0$. Then we can extend $w_1$ to a basis $\{w_1,w_2,\dots,w_n\}$ of $W$ and consider the linear form $\phi\colon W\to F$ such that $\phi(w_1)=1$ and $\phi(w_i)=0$ for $i=2,\dots,n$. Then $$ T'(\phi)(v)=\phi\circ T(v)=\phi(w_1)=1\ne0 $$ and therefore $T'(\phi)\ne0$.
Post scriptum
By definition of dual basis, if $w\in W$, you have $w=\sum_{i=1}^n \phi_i(w)w_i$. In particular, for $v\in V$, $$ Tv=\sum_{i=1}^n\phi_i(T(v))w_i $$ However, $\phi_i(T(v))=(\phi_i\circ T)(v)=\psi_i(v)$, where $\psi_i=T'(\phi_i)$. If, by assumption, $T'=0$, we have $\psi_i=0$ and so $Tv=0$. This is what you actually are using in your proof.
In my version, the basis is chosen ad hoc and so the full machinery of dual bases is not required.