If I have antilinear form $T: V \rightarrow W$ with $V,W$ finite vector spaces and $T(v_1 + a v_2 ) = T(v_1) + \bar{a} T(v_2)$ for $a \in \mathbb{C}$, I want to show the following:
(i) If $T$ is bijection, then inverse is also an antilinear map.
(ii) Rank nullity theorem holds for $T$.
For (i), if I apply $T^{-1}$ to $T(v_1 + a v_2 ) = T(v_1) + \bar{a} T(v_2)$, I get $v_1 + a v_2 = T^{-1} ( T(v_1) + \bar{a} T(v_2))$. From this it looks like the only way for left and right hand sides to be equal is it $T^{-1}$ is antilinear... but thats not really a proof.
For (ii), if I assume $dim(V) = n$, do I assume a set of basis and show different vectors span the image and kernel with total elements summing to $n$?
For (i), you need to show that $T^{-1}(w_1+aw_2) = T^{-1}(w_1) + \bar a T^{-1}(w_2)$ for all $w_1,w_2\in W$ and $a\in\mathbb C$. What you have written above is indeed useful for showing this. What happens when you set $v_i=T^{-1}(w_i)$?
For (ii), I think you have the right idea: mimic the proof of the rank-nullity theorem for linear maps.