Let $F$ be a coherent sheaf on $\mathbb{P}^1$. How to show that there exists a unique exact sequence of the form
$$0\to\mathcal{O}_{\mathbb{P}^1}(-1)^{\oplus a}\to\mathcal{O}_{\mathbb{P}^1}^{\oplus b}\to F\to\mathcal{O}_{\mathbb{P}^1}(-1)^{\oplus c}\to\mathcal{O}_{\mathbb{P}^1}^{\oplus d}\to 0?$$
There must be some more elegant arguments, but here is one.
The hypercohomology of such complex should be zero. Since $\mathcal O(-1)$ have no cohomology, the usual spectral sequence implies that $\mathcal O^{\oplus b}=H^0(F)\otimes \mathcal O$ and $\mathcal O^{\oplus d}=H^1(F)\otimes \mathcal O$. Moreover, the map $$ H^0(F)\otimes \mathcal O\to F $$ has to be the natural map. This should pretty much give uniqueness.
For the existence, let's do this. Consider the map $H^0(F)\otimes \mathcal O\to F$ and its image $G$. We have an inclusion $0\to G\to F$, which means that there is an inclusion $H^0(G)\to H^0(F)$. Since the composition $$H^0(F)=H^0(H^0(F)\otimes \mathcal O))\to H^0(G)\to H^0(F)$$ is identity, we see that $H^0(G)=H^0(F)$.
Now consider the short exact sequence $$0\to G_1\to H^0(F)\otimes \mathcal O\to G\to 0$$ where $G_1$ is the kernel. We would like to show that $G_1$ is a direct sum of some number of copies of $\mathcal O(-1)$. From the long exact sequence in cohomology we conclude that $H^0(G_1)=H^1(G_1)=0$. Since it is a sub sheaf of a locally free sheaf, it is itself locally free (we are on a curve). Moreover, on $\mathbb P^1$, it has to be a direct sum of $\mathcal O(-r_i)$, so easily all $r_i$ are equal to $1$. While we are at it, observe that the long exact sequence in cohomology shows that $H^1(G)=0$.
So we have established the first half of the sequence $$ 0\to G_1=\mathcal O(-1)^{\oplus a}\to \mathcal O^{\oplus b}\to F. $$
Now we have $$ 0\to G\to F\to G_2\to 0. $$ We know that $H^0(G)\to H^0(F)$ is an isomorphism and $H^1(G)=0$. This implies that $H^0(G_2)=0$ and $H^1(F)\to H^1(G_2)$ is an isomorphism.
Since $H^0(G_2)=0$, $G_2$ is torsion-free and is a direct sum of $\mathcal O(-r_i)$ for some $r_i>0$. So it becomes a question of showing that there is a way of writing every such bundle as a kernel of a map from some number of copies of $\mathcal O(-1)$ to some number of copies of $\mathcal O$. By dualizing, it becomes a question of writing $$ 0\to \mathcal O^{\oplus d}(-1) \to \mathcal O^{\oplus c} \to G_2^\vee(-1)\to 0. $$ As before, we see that the map $\mathcal O^{\oplus c}=H^0(G_2^\vee(-1))\otimes \mathcal O$ with the natural map. As before, the kernel can be shown to be acyclic vector bundle. We just need to argue that the map $$ H^0(G_2^\vee(-1))\otimes \mathcal O\to G_2^\vee(-1) $$ is surjective. This is immediately clear from the decomposition of $G_2$ into invertible sheaves and the fact that $\mathcal O(\geq 0)$ are globally generated.
{\bf Edit:} Here is a replacement for the spectral sequence argument. The long exact sequence splits into $$ 0\to \mathcal O(-1)^{\oplus a}\to \mathcal O^{\oplus b}\to B_1 \to 0 $$ $$ 0\to B_1\to F\to B_2 \to 0 $$ $$ 0\to B_2\to \mathcal O(-1)^{\oplus c}\to \mathcal O^{\oplus d}\to 0 $$ From the long exact sequences in cohomology, and from the knowledge of cohomology of $\mathcal O$ and $\mathcal O(-1)$, we conclude from the first and third short exact sequences that $$ H^0(B_1)=\mathbb k^{b},~H^1(B_1)=0,~H^0(B_2)=0,~H^1(B_2)=\mathbb k^d. $$ If we now apply this knowledge to $0\to B_1\to F\to B_2\to 0$, we see that $\mathbb k^b=H^0(F)$ and $\mathbb k^d=H^1(F)$. Moreover, it is clear from this argument that the map $H^0(\mathcal O^{\oplus b})\to H^0(F)$ is an isomorphism.