1. Are there simple examples (say on a curve or surface) of line bundles that are globally generated but not ample, of ample line bundles with no sections, of ample line bundles that are globally generated but not very ample, and of very ample line bundles with higher cohomology?
On a curve of genus $g$, a general divisor of degree $d \le g-1$ has no sections. Of course, if $d>0$ then it is ample.
$K_X$ on a hyperelliptic curve is globally generated but not very ample.
Look at $L=\mathcal O(1)$ on a plane curve of genus $d$. Then from
$$ 0\to \mathcal O_{\mathbb P^2}(1-d) \to \mathcal O_{\mathbb P^2}(1) \to \mathcal O_C(1)\to 0$$
you see that $H^1(\mathcal O_C(1))=H^2(\mathcal O_{\mathbb P^2}(1-d))$ which is dual to $H^0(\mathcal O_{\mathbb P^2}(d-4))$. So that's nonzero for $\ge4$.
2. Given an ample line bundle $L$, what is the minimal number $k$ so that I can be sure $L^k$ has sections, is globally generated, is very ample? Is $k$ related to the dimension of $X$?
Again, just look at the divisor of a degree 1 on a curve of genus $g$. You need $k\ge g$, so you see that there is no bound in terms of the dimension.
It turns out that a better right question to ask is about the adjoint line bundles $\omega_X\otimes L^k$ ($K_X+kL$ written additively). Then the basic guiding conjecture is by Fujita, and which says that for $k\ge \dim X+1$ the sheaf is globally generated, and for $k\ge \dim X+2$ it is very ample. This is proved for $\dim X=2$, proved with slightly worse bounds for $\dim X=3$. For higher dimensions the best result is due to Angehrn-Siu who gave a quadratic bound on $k$ instead of linear. There are some small improvements for some special cases.
3. If $L$ is very ample, I can use it to embed $X$ into some projective space. Then by projecting from points off of $X\subset \mathbb P^N$, I can eventually get a finite morphism $X\to \mathbb{P}^d$, where $d$ is the dimension of $X$. But what if I just know that $L$ is ample and globally generated? Can I also use it to get such a finite morphism to $\mathbb P^d$?
But of course $L$ gives a morphism $f$, and it follows that $f$ is finite: $f$ contacts no curve so $f$ is quasifinite, and $f$ is projective (since $X$ was assumed to be projective). And quasifinite + proper = finite.
It is not necessarily trivial in the Zariski topology. Consider for instance the plane quadric $\{x^2+sy^2+tz^2\}\subseteq \mathbb P^2\times\mathrm{Spec}\mathbb C[s,s^{-1},t,t^{-1}]$ as a family of $\mathbb P^1$'s over $\mathrm{Spec}\mathbb C[s,s^{-1},t,t^{-1}]$. It is not even isomorphic to $\mathbb P^1$ over the generic point of
$\mathrm{Spec}\mathbb C[s,s^{-1},t,t^{-1}]$ as it doesn't have a rational point. As for the multiplicativity in the Grothendieck group there are examples when it is not true not even in the localised and completed Grothendieck group, see arXiv:0903.3143.
Best Answer
I think what you want is true if $X$ and all the $Y_k$ are smooth (or have some very mild singularities e.g. quotient singularities) but I don't know many such examples. In general it appears to be false as shown by the following example:
Let $Y$ be the quadric cone given by $x_1x_2 - x_3x_4 = 0$ in $\mathbb{A}^4$. If we blow up the vertex the exceptional divisor is isomorphic to the quadric in $\mathbb{P}^3$ given by the same equation. $Y$ has a small resolution $f:X \to Y$ which is given (in the fibre over the vertex) by projecting the quadric onto one of its factors. The fibre of $f$ over the vertex is $\mathbb{P}^1$ and $f$ is a morphism of the type you want.
Since $f$ is birational and $X$ is smooth, it follows from Verdier duality that $Rf_*({\mathbb{C}}_X)$ is self dual (with a shift, depending on your conventions). However, one can see that the object in the derived category given by your formula is not self dual.