Show the bijective correspondence between
a. bijective maps $f: \mathbb{P(K^1)} \to \mathbb{P(K^1)}$ which keep the cross-ratio invariant
b. projective transformationes, i.e. $f: \mathbb{P(K^1)} \to \mathbb{P(K^1)}$ with $f([x]) = [Tx]$ for a linear transform $T: \mathbb{K^2} \to \mathbb{K^2}$
Take $f$ fullfilling the properties in b. We already showed that $f$ then keeps the cross-ratio invariant. It remains to show that f is bijective. How to procede with that? I just know that $T$ on which $f$ is based is linear, but nothing about if it is bijective.
Now take $f$ fullfilling the properties in a. Let $\det: \mathbb{K^2} \times \mathbb{K^2} \to \mathbb{K}$ be a determinant. The following hold for all points $$\mathrm{CR}(A_1,A_2,A_3,A_4) = \frac{\det(A_1,A_2)}{\det(A_2,A_3)} \frac{\det(A_3,A_4)}{\det(A_4,A_1)} = \mathrm{CR}(f(A_1),f(A_2),f(A_3),f(A_4))$$
And now? For me it seems somehow wishy-washy what my professor so far did and more appealing to our intution and hence I struggle with formally pinning down a solution to an excersice..
Second try: Let $f:\mathbb{KP^1}\to\mathbb{KP^1}$ be a bijection, which keeps cross-ratios, and let $P_1,P_2,P_3 \in \mathbb{KP^1}$ be pairwise distinct points. Since $f$ is bijective $f(P_1),f(P_2),f(P_3) \in \mathbb{KP^1}$ are also pairwise distinct points. We know that for two such triplets there exists a unique projective map $f'$, mapping $P_i$ on $f(P_i)$. Since $f'$ is projective it also keeps cross-ratios and we have $$DV\left(f(P_1),f(P_2),f(P_3),f(Q)\right) = DV\left(f'(P_1),f'(P_2),f'(P_3),f'(Q)\right) \quad \forall Q \in \mathbb{KP^1}$$ hence it follows that $f=f'$. So $f$ is a projective mapTransformation.
I am unsure about the last bit if $f$ neccessarily needs to be the same as $f'$..
Start with your second attempt, with a given bijection $f$ and a matching projective transformation $f'$ defined using $P_1,P_2,P_3$ and their images. Now consider the composition $f^{-1}\circ f'$. By construction of $f'$, it has $P_1,P_2,P_3$ as fixed points. And by the already shown implication b$\Rightarrow$a you know that it preserves cross ratios. In particular it preserves the cross ratio
$$\operatorname{DV}(P_1,P_2;P_3,X) = \operatorname{DV}(P_1,P_2;P_3,f^{-1}(f'(X)))$$
Now you can show that this cross ratio is unique: if $P_1,P_2,P_3,X$ are known, there is no other point $f^{-1}(f'(X))$ satisfying the above equation. You might use the verbose determinant expression to verify this fact. Hence
\begin{align*} f^{-1}\bigl(f'(X)\bigr) &= X \\ f'(X) &= f(X) \\ f' &= f \end{align*}