A paper of Bezrukavnikov (http://arxiv.org/abs/math/0604445) identifies this as $\mathcal{O}(-1) \oplus \mathcal{O}(-1)$ in Example 2.8.  For heuristic reasons (which I can go into if anyone cares) I believe this to be true.  Here is a partial argument I could come up with.

The identification of $G$-equivariant vector bundles on $G/B$ with $B$-representations gives us a short exact sequence of $B$-representations:
$$0 \rightarrow V_{-2} \rightarrow \mathfrak{b} \rightarrow V_0 \rightarrow 0$$
which gives us a short exact sequence in vector bundles:
$$0 \rightarrow \mathcal{O}(-2) \rightarrow \tilde{\mathfrak{sl}_2} \rightarrow \mathcal{O}(0) \rightarrow 0$$
My guess is that there is no decomposition theorem of $SL_2$-equivariant vector bundles on $\mathbb{P}^1$, so this is as good as it gets in that case.  If we are willing to pullback to the non-equivariant case, then we get a short exact sequence:
$$0 \rightarrow \mathcal{O}(-2) \rightarrow \mathcal{O}(-1) \oplus \mathcal{O}(-1) \rightarrow \mathcal{O}(0) \rightarrow 0$$
which can be written down explicitly: the nonvanishing sections on one affine cover should be $e, h \in \mathfrak{sl}_2^*$ and $f, h \in \mathfrak{sl}_2^*$ on the opposite affine cover (here I am thinking of $\mathcal{O}_{\tilde{\mathfrak{sl}_2}}$ over $\mathbb{C}[\mathfrak{sl}_2]$; the $\mathbb{C}^*$-action on the fibers may also help in thinking of those elements as being in "degree one").

It was helpful for me to think of this as a "twist" of the usual short exact sequence for the tautological bundle on $\mathbb{P}^1$:
$$0 \rightarrow \mathcal{O}(-1) \rightarrow \mathcal{O}(0) \oplus \mathcal{O}(0) \rightarrow \mathcal{O}(1) \rightarrow 0$$
though I do not know why this should be the case "geometrically" (if there is a reason at all).  I would very much appreciate it if someone could give a better explanation, or correct me if I am wrong.