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 only somewhat related 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 generating sections on one affine cover should be $e, h \in \mathfrak{sl}_2^*$ and $h, f \in \mathfrak{sl}_2^*$ (respectively!) on the opposite affine cover (here I am thinking of $\mathcal{O}_{\tilde{\mathfrak{sl}_2}}$ over $\mathbb{C}[\mathfrak{sl}_2]$, so I really mean whatever their images are in the map on rings; 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 a "geometric" reason why these two should be related (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.

EDIT: Here's one way you can prove this.  By base change on the square with $\mathbb{P}^1, \mathbb{P}^1/SL_2, *, */SL_2$ (I don't think I can input diagrams), one can take $G$-equivariant comology and then forget the $G$-action, or one can forget the $G$-action and then take cohomology. We will compute the former to deduce the latter.

Let $\mathcal{E}$ be the locally free sheaf associated to $\tilde{\mathfrak{sl}_2}$.  Take the long exact sequence of the short exact sequence of locally free sheaves and one obtains:
$$0 \rightarrow H^0(\mathbb{P}^1, \mathcal{E}) \rightarrow H^0(\mathbb{P}^1, \mathcal{O}(0)) \rightarrow H^1(\mathbb{P}^1, \mathcal{O}(-2)) \rightarrow H^1(\mathbb{P}^1, \mathcal{E}) \rightarrow 0.$$
I claim that the first map must be zero.  Above the Borel $$\left(\begin{array}{cc} * & * \\ 0 & * \end{array}\right)$$
take the point in the fiber
$$\left(\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right)$$
and use the $G$-action to try to move it around.  In particular, the action of
$$\left(\begin{array}{cc} 1 & a \\ 0 & 1 \end{array}\right)$$
fixes the Borel but moves the section to
$$\left(\begin{array}{cc} 1 & -2a \\ 0 & -1 \end{array}\right)$$
so the section does not lift and thus the first map in the short exact sequence is zero, thus the second map is an isomorphism, and the third map is zero.  So $\mathcal{E}$ has no cohomology, and now thinking of the other direction in the base-change diagram, by preservation of degree and rank, it must be $\mathcal{O}(-1) \oplus \mathcal{O}(-1)$.