Computing Holomorphic Sectional Curvatures Kobayashi in his book Hyperbolic Complex Spaces (pg. 32) defines the holomorphic sectional curvature at $p$ through $v$ on a complex manifold $(X,ds^2)$ as:
$\sup K_{f^*ds^2}(0)$
where the $\sup$ is over all holomorphic maps $f:D\to X$ with $f(0)=p$ and $\text{span}_{\mathbb{C}}f'(0)=\text{span}_{\mathbb{C}}v$ ($D$ is the unit disk, $K$ is the Gaussian curvature on $(D, f^*ds^2)$).
I have gotten confused trying to verify this formula in the following case (to compute the curvature of Fubini-Study metric):
$\mathbb{C}^2\backslash 0$ with metric $\frac{1}{|z_1|^2+|z_2|^2}ds_{eucl}^2=ds^2$ and sectional curvatures $\overline K$.
We should have $\overline K_p(v,Jv)=1$ for $v\in \mathbb{C} p^\perp$.
However when I compute with Kobayashi's definition I find $\overline K_p(v,Jv)=2$ as follows:
$f^*ds^2=\frac{|f'(z)|^2}{|f(z)|^2}dzd\overline z$, and using
$K_{\lambda dzd\overline z}=-2\frac{\partial \overline\partial\log\lambda}{\lambda}$ I get:
$K_{f^*ds^2}(0)=2(1-\frac{|\langle p, v\rangle|^2}{|p|^2|v|^2}+\frac{|p|^2}{|v|^6}(|\langle v, f''(0)\rangle|^2-|v|^2|f''(0)|^2)\le 2(1-\frac{|\langle p, v\rangle|^2}{|p|^2|v|^2}))$
by Cauchy-Schwarz and it is realized.
So I am off by a multiple of 2! I cannot find my misunderstanding of how did I get this extra multiple?
 A: I believe this is the fix:
Call Kobayashi's holomorphic sectional curvature the one by
$H(X)=\sup K_{f^*ds^2}(0)$ with $f'(0)\in \mathbb{C}X$.
Call the holomorphic sectional curvature of $ds^2$ the sectional curvature through a complex plane:
$K^{hol}(v)=\overline K(v,Jv)$.
These are not always the same. Consider conformally flat metric: $g=\lambda\sum  (dx^i)^2+(dy^i)^2$. With associated hermitian metric $g^\mathbb{C}$.
Kobayashi proves for $X=X^i\frac{\partial}{\partial z^i}$ a unit vector of $g^\mathbb{C}$ that:
$H(X)=R_{i\overline j k\overline l} X^i\overline X^j X^k\overline X^l$
where $R_{i\overline j k\overline l}=-\frac{\partial}{\partial z^k}\frac{\partial}{\partial \overline z^l} g_{i\overline j}+g^{p\overline q}\frac{\partial}{\partial z^k} g_{i\overline q}\frac{\partial}{\partial \overline z^l}g_{p\overline j}$
Taking wlog $X=\sqrt{\frac{2}{\lambda}}\frac{\partial}{\partial z^1}$ (note $g^\mathbb{C}(\frac{\partial}{\partial z^1}, \frac{\partial}{\partial \overline z^1})=\lambda/2$) and $v=\sqrt{\frac{1}{\lambda}}\frac{\partial}{\partial x^1}$.
Then I compute:
$H(X)=4\frac{R_{1\overline 1 1 \overline 1}}{\lambda^2}=\frac{R_{1212}}{\lambda^2}+\frac{\sum_{i\ge 2} \frac{\partial}{\partial z^i}\lambda \frac{\partial}{\partial \overline z^i}\lambda}{\lambda^3}=K^{hol}(v)+\frac{\sum_{i\ge 2} \frac{\partial}{\partial z^i}\lambda \frac{\partial}{\partial \overline z^i}\lambda}{\lambda^3}.$
Back to the example with $\lambda=(z^1\overline z^1+z^2\overline z^2)^{-1}$  we take $X=\sqrt{\frac{2}{\lambda}}\frac{\partial}{\partial z^1}$ so that $X\in \mathbb{C} q^\perp$ makes $q=(0,z^2)$.
Now we compute:
$H(X)=2$ (as computed in the question)
$\frac{\frac{\partial}{\partial z^2}\lambda \frac{\partial}{\partial \overline z^2}\lambda}{\lambda^3}=z^2\overline z^2\lambda$ which is 1 when evaluated at $q$.
So
$\overline K= K^{hol}(\frac{\partial}{\partial x^1})=1$ (this is what it should be to agree with the Fubini-Study after pushing down via submersion)
