Let $\mathbb{C}P^2=(\mathbb{C}^3\setminus\{0\})/\mathbb{C}^\times$ be the complex projective plane. Let
$E=\{([z],x)\in\mathbb{C}P^2\times\mathbb{C}^3\mid x\in[z]\}$
be the universal line bundle and let
$E^\perp=\{([z],y)\in\mathbb{C}P^2\times\mathbb{C}^3\mid \langle y,z\rangle=0\}$
be its complementary complex $2$-plane bundle, where the angular brackets denote the standard Hermitian form on $\mathbb{C}^3$ that is complex linear in the second variable. These are both subbundles of the trivial bundle $\mathbb{C}P^2\times\mathbb{C}^3$ and satisfy $E\oplus E^\perp\cong\mathbb{C}P^2\times\mathbb{C}^3$.
Then there is a well-known isomorphism of vector bundles
$$T\mathbb{C}P^2\cong \operatorname{Hom}(E,E^\perp)\cong E^*\otimes E^\perp$$
where $T\mathbb{C}P^2$ is the tangent bundle of $\mathbb{C}P^2$ and $E^*=\operatorname{Hom}(E,\mathbb{C}P^2\times\mathbb{C})$ is the dual bundle. See for instance Milnor and Stasheff's Characteristic Classes, pg. 169. Everything generalises by working over $\mathbb{C}P^n$, but for simplicity I'll stick to the $n=2$ case.
I've left my statement above a little vague because it is my first question:
What is $T\mathbb{C}P^2$ in the isomorphism above? Is it the rank 4 real tangent bundle of the 4-manifold $\mathbb{C}P^2$, and the isomorphism is forgetting the complex structure on the hom-bundle? Or is it the complex tangent bundle of the complex surface $\mathbb{C}P^2$?
My difficulty arises because I have calculated the transition functions of $E$ and $E^\perp$, and the product $E^*\otimes E^\perp$ does not seem to be a holomorphic bundle. In fact, on the level of transition functions I am having difficulty matching them up in any way, which is the aim of the exercise I am attempting. The details follow.
Everything is trivialisable over the open subsets $U_i=\{[z_0,z_1,z_2]\mid z_i\neq 0\}\subseteq\mathbb{C}P^2$, $i=0,1,2$ which are manifold charts on $\mathbb{C}P^2$ with homeomorphisms $\varphi_i:\mathbb{C}^2\xrightarrow{\cong}U_i$. For ease of exposition I'll focus on the the transitions over $U_0\cap U_1$. I find trivialisations $\theta_i:U_i\times \mathbb{C}\rightarrow E^*|_{U_i}$ and $\psi_i:U_i\times\mathbb{C}^2\rightarrow E^\perp|_{U_i}$ and calculuate the transition functions $s:U_0\cap U_1\rightarrow \mathbb{C}^\times$ induced by $\theta_1^{-1}\circ\theta_0$ and $t:U_0\cap U_1\rightarrow Gl_2(\mathbb{C})$ induced by $\psi_1^{-1}\circ\psi_0$, finding
$$s([z_0,z_1,z_2])=\frac{z_0}{z_1},\qquad t([z_0,z_1,z_2])=\begin{pmatrix}-\frac{\bar z_1}{\bar z_0}&-\frac{\bar z_2}{\bar z_0}\\0&1\end{pmatrix}.$$
The corresponding transition function for $E^*\otimes E^\perp$ should therefore be
$$(s\otimes t)([z_0,z_1,z_2])=\begin{pmatrix}-\frac{\bar z_1 z_0}{\bar z_0 z_1}&-\frac{\bar z_2 z_0}{\bar z_0 z_1}\\0&\frac{z_0}{z_1}\end{pmatrix}.$$
On the other hand, the transition function $u:U_0\cap U_1\rightarrow Gl_2(\mathbb{C})$ for the holomorphic tangent bundle is given by the Jacobian of the cordinate transformation $\varphi_1^{-1}\circ \varphi_0(x,y)=(1/x,y/x)$, which we evaluate at $\varphi_0^{-1}([z_0,z_1,z_2])$ to match the above. We find
$$u([z_0,z_1,z_2])=\begin{pmatrix}-\frac{z_0^2}{z_1^2}&0\\-\frac{z_2z_0}{z_1^2}&\frac{z_0}{z_1}\end{pmatrix}.$$
Firstly I cannot see how these transition functions are equivalent, which would mean that there would be smooth functions $h_i:U_i\rightarrow Gl_2(\mathbb{C})$ such that $u=h_1^{-1}\cdot (s\otimes t)\cdot h_0$. Secondly, it doesn't seem like $(s\otimes t)$ is even holomorphic.
Where am I going wrong in my calculations? What am I not understanding?