For concrete 2-dimensional counter-example see page 328 of Kulkarni's paper "Curvature and metric", Annals of Math, 1970. Weinstein's argument there shows that every Riemannian surface provides a counter-example (using flow orthogonal to the gradient of the curvature).
On the positive side, if $M$ is compact of dimension $\ge 3$ and has nowhere constant sectional curvature, then combination of results of Kulkarni and Yau show that
a diffeomorphism preserving sectional curvature is necessarily an isometry.
Concerning 2-dimensional counter-examples: First of all, every surface which admits an open subset where curvature is (nonzero) constant would obviously yield a counter-example. Thus, I will assume now that curvature is nowhere constant. Kulkarni refers to Kreyszig's "Introduction to Differential Geometry and Riemannian Geometry", p. 164, for a counter-example attributed to Stackel and Wangerin. You probably can get the book through interlibrary loan if you are in the US. Here is the Weinstein's argument. Consider a Riemannian surface $(S,g)$ and let $K$ denote the curvature function. Let $X$ be a nonzero vector field on $S$ orthogonal to the gradient field of $K$. (Such $X$ always exists.) Now, consider the flow $F_t$ along $X$. Then $F_t$ will preserve the curvature but will not be (for most metrics $g$ and vector fields $X$) isometric. For instance, $F_t$ cannot be isometric if genus of $S$ is at least $2$ or if $X$ has more than 2 zeros.
Since you seem to have some basic misunderstandings about these concepts, I will try to clarify the definitions.
Let $(M,J)$ be a complex manifold, with a Riemannian metric $g$ which is Hermitian, i.e. $g(JX,JY)=g(X,Y)$ for all $X,Y\in TM$ real tangent vectors. Assume furthermore that $g$ is Kähler, i.e. that the real $2$-form $\omega(X,Y)=g(JX,Y)$ is closed.
Then it is well-known that the Riemann curvature tensor of $g$ satisfies the "Kähler identity"
$$R(X,Y,Z,W)=R(X,Y,JZ,JW).$$
We will use the convention that if $X,Y$ are nonparallel unit vectors, then the sectional curvature of the $2$-plane they span is $R(X,Y,Y,X)$.
Then for a unit vector $X$ we define its holomorphic sectional curvature $H(X)$ to be the sectional curvature of the $2$-plane spanned by $X$ and $JX$, i.e.
$$H(X)=R(X,JX,JX,X).$$
In analogy with this, we define the bisectional curvature of two unit vectors $X,Y$ to be
$$R(X,JX,JY,Y).$$
The name comes from the fact that (using the Bianchi and Kähler identities)
$$R(X,JX,JY,Y)=R(X,Y,Y,X)+R(X,JY,JY,X),$$
so that if $X$ and $Y$ are not parallel it is indeed equal to the sum of two sectional curvatures (incidentally, this might be the relationship you wanted between sectional and bisectional curvature).
Then every simply connected Kähler manifold with constant holomorphic sectional curvature is holomorphically isometric to one of the three model spaces $\mathbb{C}^n, B^n$ and $\mathbb{CP}^n$ with (a rescaling of) their standard metrics. This is proved for example in Kobayashi-Nomizu "Foundations of Differential Geometry, Vol.2", Theorem 7.9.
In fact, by Proposition 7.3 in that same book, $H(X)$ is constant equal to $c$ iff
$$R(X,Y,Z,W)=\frac{c}{4}\bigg(g(X,W)g(Y,Z)-g(X,Z)g(Y,W)+g(X,JW)g(Y,JZ)$$
$$-g(X,JZ)g(Y,JW)+2g(X,JY)g(W,JZ)\bigg).$$
In particular, in this case you have that the bisectional curvature of $X,Y$ equals
$$R(X,JX,JY,Y)=\frac{c}{2}\bigg(1+g(X,Y)^2+g(X,JY)^2\bigg).$$
As you can see, this expression is not constant (as you vary $X,Y$ among unit vectors), but it varies between $\frac{c}{2}$ and $c$ according to the relative position of the $2$-planes spanned by $(X,JX)$ and $(Y,JY)$.
In particular you see that there is no such thing as a non-flat "Kähler manifold with constant bisectional curvature", because if there was such a thing, in particular the holomorphic sectional curvature would be constant, but then the curvature tensor would be given by the above formula, and the bisectional curvature would actually be NOT constant. On the other hand, if $c=0$, then all the curvatures are constant because the metric is flat.
It goes without saying that Tian's book has a recurrent typo, and wherever he says "constant bisectional curvature" you should substitute "constant holomorphic sectional curvature".
Apart from the book of Kobayashi-Nomizu, you can also read this paper of Goldberg-Kobayashi.
Best Answer
I think the idea is to think of $\hat{R}_p$ as a mapping from $\Lambda^pM$ to itself, and $\hat{\omega}$ as a mapping from $\Lambda^pM$ to $E$ (the vector bundle in which $\omega$ takes values), and then just compose these two operators (to get something $E$ valued in the end).
In the case where $E$ is the trivial (scalar) bundle, it works out to be $\hat{R}_p(\omega)$ since $R_p$ is symmetric.