There is a more enlightening proof of this statement than Berger's calculation and, in fact, it proves something a bit more general. First, a definition: Let $(M,g)$ be a Riemannian $n$-manifold with Riemann curvature tensor $R$ and let $E\subset T_xM$ be a $p$-plane with orthonormal basis $e_1,\ldots,e_p$. Define
$$
\sigma(E) = \sum_{i,j=1}^p R(e_i,e_j,e_j,e_i),
$$
which is easily seen not to depend on the choice of orthonormal basis.
The quantity $\sigma(E)$ is the scalar curvature of $E$. For $p=2$, this is twice the sectional curvature of the $2$-plane $E$, while $\sigma(T_xM)$ is simply the scalar curvature of $g$ at $x$. In general, the scalar curvature of $E$ is $p(p{-}1)$ times the average of the sectional curvatures of the $2$-planes that lie in $E$.
Fact: If $(M,g,J)$ is a Kähler manifold, and $E\subset T_xM$ is a $4$-plane that is complex (i.e., stable under $J$), then $\sigma(E)$ is $6$ times the average sectional curvature of the complex $2$-planes that lie in $E$. (See below for a proof.)
Proposition: If $(M,g,J)$ is a Kahler manifold and $E\subset T_xM$ is a $4$-plane that is complex and satisfies $\sigma(E)>0$, then the only local maxima of the sectional curvature function on $\mathrm{Gr}_2(E)$ occur at $2$-planes $L$ that are complex.
Remark: If the holomorphic sectional curvatures of $M$ are positive, then, by the Fact, $\sigma(E)>0$ for all $4$-planes $E$ that are complex, and hence, by the Proposition, the maximum of the sectional curvature on $\mathrm{Gr}_2(T_xM)$ is attained only by complex $2$-planes (since each noncomplex tangent $2$-plane lies in some tangent $4$-plane that is complex). Meanwhile, one can have $\sigma(E)>0$ for all $4$-planes that are complex without having positive holomorphic sectional curvature; just consider a Kähler surface with positive scalar curvature that does not have positive holomorphic sectional curvature. Thus, this gives a stronger result than Klingenberg's claim.
To prove the Fact and the Proposition, it clearly suffices to do it for a Kähler manifold of complex dimension $2$ with positive scalar curvature, in which case $E=T_xM$ for some $x\in M$ that one can suppose fixed for the purposes of this discussion. This is a local argument, so one can choose an orthonormal coframing $\omega_0,\ldots,\omega_3$ such that the Kähler form is given by $\Omega=\omega_0\wedge\omega_1+\omega_2\wedge\omega_3$. Let $\phi_{ij}=-\phi_{ji}$ be the connection forms, so that $d\omega_i=-\phi_{ij}\wedge\omega_j$ and, because of the Kähler condition, $\phi_{02}+\phi_{31}=\phi_{03}+\phi_{12}=0$. Then, by the Bianchi identities for a Kahler metric, one has the following expression for the curvature forms $\Phi_{ij}=d\phi_{ij}+\phi_{ik}\wedge\phi_{kj}\ $:
$$
\begin{pmatrix}
\Phi_{01}+\Phi_{23}\cr
\Phi_{02}+\Phi_{31}\cr
\Phi_{03}+\Phi_{12}\cr
\Phi_{01}-\Phi_{23}\cr
\Phi_{02}-\Phi_{31}\cr
\Phi_{03}-\Phi_{12}
\end{pmatrix} =
\begin{pmatrix}
3s&0&0&r_1&r_2&r_3\cr
0&0&0&0&0&0\cr
0&0&0&0&0&0\cr
r_1&0&0&s{+}w_{11}&w_{12}&w_{13}\cr
r_2&0&0&w_{12}&s{+}w_{22}&w_{23}\cr
r_3&0&0&w_{13}&w_{23}&s{+}w_{33}
\end{pmatrix}
\begin{pmatrix}
\omega_0\wedge\omega_1+\omega_2\wedge\omega_3\cr
\omega_0\wedge\omega_2+\omega_3\wedge\omega_1\cr
\omega_0\wedge\omega_3+\omega_1\wedge\omega_2\cr
\omega_0\wedge\omega_1-\omega_2\wedge\omega_3\cr
\omega_0\wedge\omega_2-\omega_3\wedge\omega_1\cr
\omega_0\wedge\omega_3-\omega_1\wedge\omega_2
\end{pmatrix}
$$
Here, $12s$ is the scalar curvature (which is positive by hypothesis) and $w_{ij}=w_{ji}$ satisfies $w_{11}{+}w_{22}{+}w_{33}=0$. [The reader may recognize this as the classic presentation of the curvature of a Riemannian $4$-manifold due to Singer and Thorpe; in this case, it has been adapted to the case of a Kähler metric.]
Define functions $u_i$ and $v_i$ on the Grassmann bundle $\mathrm{Gr}^+_2(M)$ of oriented tangent $2$-planes as follows: If $L$ is an oriented $2$-plane with oriented orthonormal basis $(X,Y)$, then
$$
\begin{pmatrix}
u_1(L)\cr
u_2(L)\cr
u_3(L)\cr
v_1(L)\cr
v_2(L)\cr
v_3(L)
\end{pmatrix} =
\begin{pmatrix}
\omega_0\wedge\omega_1+\omega_2\wedge\omega_3\cr
\omega_0\wedge\omega_2+\omega_3\wedge\omega_1\cr
\omega_0\wedge\omega_3+\omega_1\wedge\omega_2\cr
\omega_0\wedge\omega_1-\omega_2\wedge\omega_3\cr
\omega_0\wedge\omega_2-\omega_3\wedge\omega_1\cr
\omega_0\wedge\omega_3-\omega_1\wedge\omega_2
\end{pmatrix}
(X,Y).
$$
One then has ${u_1}^2+{u_2}^2+{u_3}^2={v_1}^2+{v_2}^2+{v_3}^2=1$, and the map $(u,v):\mathrm{Gr}^+_2(M)\to S^2\times S^2$ carries each fiber of $\mathrm{Gr}^+_2(M)\to M$ diffeomorphically onto $S^2\times S^2$. Note that a $2$-plane $L$ is complex if and only if $u_2(L)=u_3(L)=0$, i.e., if and only if $u_1(L)=\pm1$.
Now, computing from the definitions, one finds
$$
\sigma(L) = 3s\ {u_1}^2 + 2u_1(r_iv_i) + s + w_{ij}v_iv_j\ .
$$
(The proof of the Fact about the average of sectional curvatures of complex $2$-planes in $E$ follow immediately from this formula.)
Now, $\sigma(L)$ does not depend on $u_2$ or $u_3$, and we can thus think of $\sigma$ restricted to a single fiber $\mathrm{Gr}^+_2(T_xM)$ of $\mathrm{Gr}^+_2(M)\to M$ as a function on $[-1,1]\times S^2$ instead of $S^2\times S^2$. Because $s>0$, when one fixes a $v\in S^2$, the quadratic function of $u_1$ above cannot attain a local maximum when $-1 < u_1 <1 $. It follows that the local maxima of $\sigma$ on $\mathrm{Gr}^+_2(T_xM)$ must occur only at those $L$ for which $u_1(L)=\pm1$, i.e., those $L$ that are complex. QED
