For a $v\in(\ker T)^\perp$, the linear form $\phi\in V^*$ given by $\phi(x)=\langle x,v\rangle$ vanishes on $\ker T$. This means it passes to the quotient: there is a well defined linear form $\phi'\in(V/\ker T)^*$ such that $\phi'(\overline v)=\phi(v)$ for all $v\in V$, where $\overline v=v+\ker T$ is the class of $v$ in the quotient. By the first isomorphism theorem, $V/\ker T$ is isomorphic to $\operatorname{Im}T$ via the map induced by $T$, so $\phi'$ gives a linear form on $\operatorname{Im}T\subseteq V$, which can be extended to a linear form $\psi$ on all of $V$; by construction it satisfies $\psi(T(x))=\phi(x)$ for all $x\in V$. Now if $w\in V$ is the unique vector such that $\psi(y)=\langle y,w\rangle$ for all $y\in V$, then for all $x\in V$ one has
$$
\langle x,T^*(w)\rangle = \langle T(x),w\rangle =\psi(T(x))= \phi(x)=\langle x,v\rangle
$$
and it follows that $v=T^*(w)$, giving $(\ker T)^\perp\subseteq \operatorname{Im}(T^*)$ as desired.
Alternatively, since we are in finite dimension, you could use the rank-nullity theorem to conclude that $\dim((\ker T)^\perp)=\operatorname{rk} T=\operatorname{rk} T^*=\dim \operatorname{Im}(T^*)$, so that the inclusion you already had suffices. But the above aregument lets you actually find $w$ mapping to $v$; there is a choice involved, that comes from the choice of the extension $\psi$ from $\operatorname{Im}T$ to $V$ of a given linear form.
