Since the problem has a local nature, I'll consider only the Linear Algebra couterpart. Let $V$ be a real vector space whose dimension is $2n$ and let $$J\colon V \longrightarrow V$$ be a complex structure on $V$, $J^2 = -Id$. Then $(V,J)$ is a (n-dimensional) complex vector space for the scalar multiplication $$(a+bi)v = av+bJ(v).$$ Let $\left<,\right>$ be a (real) inner product on $V$ compatible with $J$, $$\left<J(v),J(w)\right> = \left<v,w\right>.$$ We define an hermitian product on $(V,J)$: $$\left(v,w\right) := \left<v,w\right> -i\left<J(v),w\right>.$$ On the other hand, we have a natural hermitian extension for $\left<,\right>$ to $V\otimes\mathbb{C}$: $$\left<v\otimes a,w \otimes b\right>_{\mathbb{C}} := a\bar b \left<v,w\right>.$$
Extending $J$ to $V\otimes\mathbb{C}$ we may take the eingenspaces for $J$ and make the decomposition $$V\otimes\mathbb{C} = V'\oplus V''.$$ Fix a (real) $\mathbb{C}-$basis $\{x_1, \dots x_n\}$ for $(V,J)$. Then the $y_j := J( x_j)$ complete a real basis for $V$. Define $$z_j = \frac{1}{2}(x_j -iy_j).$$ Since $J(z_j) =iz_j$, the set $\{z_1, \dots z_n\}$ is a $\mathbb{C}-$basis for $(V',i)$, the set $\{\bar z_1, \dots \bar z_n\}$ is a $\mathbb{C}-$basis for $(V'',-i)$ and $x_j \mapsto z_j$ , $x_j \mapsto \bar z_j$ are isomorphisms. Now we head to the product.
$$ \left<z_j,z_k\right>_{\mathbb{C}} = \frac{1}{4}\left[\left<x_j,x_k\right> -i\left<y_j,x_k\right> +i \left<x_j,y_k\right> +\left<y_j,y_k\right> \right] = \frac{1}{2}\left(x_j, x_k\right),
$$
$$ \left<z_j,\bar z_k\right>_{\mathbb{C}} = \frac{1}{4}\left[\left<x_j,x_k\right> -i\left<y_j,x_k\right> -i \left<x_j,y_k\right> -\left<y_j,y_k\right> \right] = 0,
$$
$$ \left<\bar z_j,\bar z_k\right>_{\mathbb{C}} = \overline{\left<z_j,z_k\right>}_{\mathbb{C}}.
$$
We have then the relation between the hermitian products $\left(,\right)$ on $(V,J)$ and $\left<,\right>_{\mathbb{C}}$ on $(V\otimes \mathbb{C},i)$. The last thing to remark is that on Kobayashi's notation $$ h(z_j,z_k) = \left<z_j,\bar z_k\right>_{\mathbb{C}}.$$
Best Answer
Note that the formal definition of the pullback of $\pi : L \to Y$ by $f : X \to Y$ is as follows: the total space is given by $f^*L = \{(x, \ell) \in X\times L \mid f(x) = \pi(\ell)\}$ and the projection $\pi' : f^*L \to X$ is given by projection onto the first factor; i.e. $\pi'(x, \ell) = x$. There is also a map $F : f^*L \to L$ given by projection onto the second factor, i.e. $F(x, \ell) = \ell$. These maps give rise to a commutative square:
$$\require{AMScd} \begin{CD} f^*L @>{F}>> L\\ @V{\pi'}VV @VV{\pi}V \\ X @>{f}>> Y \end{CD}$$
That is, $\pi\circ F = f\circ\pi'$.
Note that for $x_0 \in X$, we have $$(f^*L)_{x_0} = \pi'^{-1}(x_0) = \{(x_0, \ell) \in X\times L \mid f(x_0) = \pi(\ell)\} \cong \pi^{-1}(f(x_0)) = L_{f(x_0)}.$$
The pullback metric $f^*h$ on $(f^*L)_{x_0}$ is just the original metric $h$ on $L_{f(x_0)}$. More precisely, $$(f^*L)_x(s(x), t(x)) = h_{f(x)}(F(s(x)), F(t(x))).$$