Image of dual map is annihilator of kernel

By the first isomorphism theorem the map $T$ factors over $V/\ker T$ as $$V\ \stackrel{\pi}{\longrightarrow}\ V/\ker T\ \stackrel{\overline{T}}{\longrightarrow}\ \operatorname{im}T,$$ where $\pi$ is the canonical quotient map and $\overline{T}$ is an isomorphism. Note that $\pi=\overline{T}^{-1}\circ T$.

If $\phi\in(\ker T)^{\circ}$ then $\ker T\subseteq\ker\phi$. This means $\phi$ also factors over $V/\ker T$, as $$V\ \stackrel{\pi}{\longrightarrow}\ V/\ker T\ \stackrel{\psi}{\longrightarrow}\ F,$$ where $F$ denotes the base field, and $\psi\in(V/\ker T)'$. It follows that $$\phi=\psi\circ\pi=\psi\circ(\overline{T}^{-1}\circ T)=(\psi\circ\overline{T}^{-1})\circ T,$$ where of course $\overline{T}^{-1}$ is an isomorphism, hence $\psi\circ\overline{T}^{-1}\in(\operatorname{im}T)'$. Because $\operatorname{im}T$ is a linear subspace of $W$, the linear functional $\psi\circ\overline{T}^{-1}$ on $\operatorname{im}T$ extends to a linear functional $f$ on $W$, which then satisfies $$\phi=f\circ T,$$ by construction, as desired. Note that $f$ is far from unique in general.