In $SU(n+1)/U(n)$ there is a natural basepoint, the coset of the identity, which is fixed by the action of $U(n)$ (thought of as acting on the quotient by virtue of being a subgroup of $SU(n1)$). Since $U(n)$ fixed this point, it acts on the (complexified) cotangent space to this point, and the problem is then to understand this representation, and in particular, to decompose it into two pieces, the $(1,0)$ piece and the $(0,1)$ piece.
Now the tangent space is ${\mathfrak su}(n+1)/{\mathfrak u(n)}$ (here I mean complexified Lie algebras), and so we have to decompose
this quotient under the adjoint action of $U(n)$. It has dimension $(n+1)^2-1-n^2 = 2n,$
and in fact it will decompose as the sum of the standard representation of $U(n)$ direct sum
its dual (or equivalently, its complex conjugate). One of these representations will
give the $(1,0)$-subbundle of the tangent bundle, and the other the $(0,1)$-subbundle.
Dualizing (to pass from tangent to cotanget) will give your answer. (I am not going
to actually stipulate which is which, just because I'm likely to blunder while tracing through the constructions and the duality; but it shouldn't
be hard to work out if you sit down with pen and paper.)
[The following is added in response to the comment below, asking for the movitation behind the above calculation; hopefully it is of some help:]
Since we are looking for an $SU(n+1)$-equivariant splitting of the (co)tangent bundle, it is enough to look for a $U(n)$-equivariant splitting of its fibre at the $U(n)$-fixed point. (This is a manifestation of the very reason that $U(n)$-reps. give rise to equivariant bundles on the quotient.) At that fixed point (the identity coset) the tangent space is ${\mathfrak su(n+1)}/{\mathfrak u(n)}$, just because it is the quotient of the tangent spaces at the identity of the corresponding groups. Since this has to split into two complex conjugate halves under the $U(n)$ action (we know that is breaks up into a $(1,0)$ and $(0,1)$ part), each of dimension $n$, it's not hard to guess what they must be. A little computation in the Lie algebra ${\mathfrak su(n+1)}$ confirms the guess.
Maybe a general lesson to be drawn is: when trying to compute a $G$-equivariant bundle on $G/H$, it is enough to compute the $H$-representation on the fibre at the identity coset; indeed, passing from $G$-equivariant bundles to this fibre is the quasi-inverse functor to the one (implicitly) alluded to in the introduction, which associates a $G$-equivariant bundle to an $H$-representation.