In Jack's example the fiber is not scheme-theoretically $\mathbb A^1$. You can get a counterexample by taking $Y$ to be a nodal curve, $Y'$ its the normalization, with one of the two points in the inverse image of the node removed, and $X = Y' \times \mathbb A^1$.
If we assume that the map is smooth, this becomes quite subtle. It is false in positive characteristic. Let $k$ be a field of characteristic $p > 0$. Take $Y = \mathbb A^1 = \mathop{\rm Spec}k[t]$, $Y' = \mathop{\rm Spec} k[t,x]/(x^p - t)$. Of course $Y' \simeq \mathbb A^1$, but the natural map $Y' \to Y$ is an inseparable homemorphism. Now embed $Y'$ in $\mathbb P^1 \times Y$ over $Y$, and take $X$ to be the difference $(\mathbb P^1 \times Y) \smallsetminus Y'$.
On the other hand, it is not so hard to show that in characteristic 0 the answer is positive for $n = 1$ (if $Y$ is reduced), and I believe it is known to be true $n = 2$. The general case seems estremely hard.
I am afraid that Sasha'a argument does not work; if the fiber does not have a vector spaces structure, there is not reason that choosing points gives you a trivialization.
[Edit] The question has been updated with "what if one assumes vector space structure on the fibres?"?
Well, $\mathbb A^n$ can always be given a vector space structure. In my first example, the fibers are canonically isomorphic to $\mathbb A^1$, so they have a natural vector space structure.
However, if the map $X \to Y$ is smooth, and the vector space stucture is allowed to "vary algebraically" that is, if the zero section $Y \to X$ is a regular function, the addition gives a regular function $X \times_Y X \to X$, and scalar multiplication gives a regular function $\mathbb A^1 \times X \to X$, then $X$ is in fact a vector bundle. The proof uses some machinery: one uses smoothness to construct bases locally in the étale topology, showing that $X$ is étale locally trivial over $Y$, and descent theory to show that in fact $X$ is Zariski locally trivial.
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.
Best Answer
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.