What leads you to believe they must accelerate? I would argue that they don't. I would also bet that your confusion comes from taking the analogy of incompressible fluid through pipes to the flow of electricity a bit too far.
When electrons (quasiparticles) move through the conduction band of a conductor (semiconductor) you cannot view this as some fluid moving through a pipe and have a complete analogy. For the case of the fluid, it will speed up only if the cross section of the pipe right before the narrow part is 'full' for the fluid flow current to be constant. The cross section of a wire is not 'full' like this in the case of electrical conduction.
Charged particles will accelerate when another force is applied, hence maybe if some segment of wire that has a different potential applied across it one would see an acceleration. I would say that for a single wire connected to two terminals that you should not see this. If the wire got too thin, the resistance would be too great and you would see a change in current.
Side note: acceleration can occur with either $\pm$ sign, do not use the word deceleration. This is something the public (and new physics students) tends to get wrong and can be confusing for people learning physics.
Your confusion stems from a fundamental misunderstanding about drift velocity. Drift velocity is not the average speed of electron motion, but instead is the average velocity vector. The average speed of free electron motion in a metal can be approximated to be the Fermi speed
$$v_F = \sqrt{\frac{2E_F}{m_e}}$$
where $E_F$ is the Fermi energy. This is incredibly fast - inserting $E_F=10$ eV gives a result that is well over $1000$ km/s.
These electrons are traveling in a solid, though, which is rife with objects to collide with, including other electrons. Therefore, the mean free path of electrons in a metal (i.e. the distance an electron travels until it collides) is typically less than $1$ nm. Therefore, these electrons almost instantaneously collide with something else. A large number of these collisions would serve to essentially randomize the direction of travel of any given electron. When you add a bunch of uniformly-randomly-distributed vectors of roughly equal length together, the resultant is essentially zero, regardless of the actual length of the vectors you added. Therefore, the average velocity vector of an electron should be close to zero, and certainly should be much smaller than its average speed, since its velocity is pointed in an essentially random direction.
When an electric field is applied to a metal, it accelerates electrons in a certain direction, and therefore alters the probability distribution of electron velocity. Velocities in the direction of the field become less probable, and velocities against the direction of the field become more probable. The longer the electric field is allowed to act on a freely-moving electron, the more this probability distribution is distorted. But, as was previously discussed, the time between collisions is quite small due to the density of metals. This means that the electric field can only alter the velocity distribution slightly, which shifts the average velocity vector (i.e. the drift velocity) slightly away from zero.
Another misunderstanding arises from the false assumption that the speed of an electrical signal in a metal is equal to the either the Fermi speed or the drift velocity. In reality, it is unrelated to either of those things. Instead, for conductors the speed of an electrical signal is given by the group velocity of an electromagnetic wave:
$$v_g=\frac{d\omega}{dk}$$
where $\omega(k)$ is the dispersion relation, and is in general derived from the band structure of the material in question. For a good (i.e. close-to-ideal) conductor, the dispersion relation is
$$\omega(k)=\frac{2k^2}{\mu\sigma}$$
for a material with conductivity $\sigma$ and permeability $\mu$. Then the group velocity is
$$v_g=\sqrt{\frac{8\omega}{\mu\sigma}}$$
which, for copper, with $\sigma= 5.96\times 10^7$ S/m and $\mu\approx\mu_0=4\pi\times 10^{-7}$ H/m, and for a plane wave with frequency $1$ GHz, the group velocity is roughly $25$ km/s, and increases with increasing frequency.
EDIT:
The conductivity of a material $\sigma$ is defined by
$$\mathbf{J}=\sigma \mathbf{E}$$
for current density $\mathbf{J}$ and applied electric field $\mathbf{E}$. This essentially means that it's the average number of electrons passing through a unit area per unit time, per unit applied electric field. The higher the conductivity, the less electric field it takes to get electrons to flow. One simple model (specifically, the Drude model) based on similar arguments as above finds that for a material with electron density $n$ and mean time between collisions $\tau$, for DC currents one has
$$\sigma = \frac{ne^2\tau}{m_e}$$
Resistivity $(\rho)$ is defined as the inverse of conductivity. Therefore, again from the Drude model for DC currents, one has
$$\rho = \frac{m_e}{ne^2\tau}.$$
Best Answer
Edit after rereading the quesiton.
Agree that it is (all else being equal) easier for electrons to travel down a wire with a greater cross sectional area. (Not sure why this is not intuitive - for me it is easier to walk down a wide pavement than a narrow passage between two buildings - particularly if other people are about to provide some 'resistance' - also it is perhaps a bit like putting two resistors in parallel reduces the resistance - e.g. for two resistors of $R$ resistance in parallel the combined resistance is ${1 \over 2} R$)
Agree that thinner wires have higher electric fields (if compared to thicker wires with the same resistivity). Your analysis is correct if resistivity is constant.
One factor that you have not considered that might be important with some resistors is that the average speed of electrons will also depend on the number of free electrons per unit volume. This density of free electrons will contribute to the resistivity, but it is not the only factor that determines resistivity. In fact, another way to do the analysis in your question would be to think about the number of free electrons passing a point in unit time - for a narrower wire with less electrons per unit length of wire the speed has to be higher, but really this is just restating what you have in your question.
(from original answer)
the exact electron speed is harder to pin down. It is more accurate to think of 'average drift velocity' because the electrons are moving about so fast in the metal in all directions all the time that applying a voltage just slightly shifts the average speed over a bit in one direction so that there is a relatively small net drift velocity, which is the current we can measure. (Here small is used because teh net drift velocity will be very much smaller than the mean speed of free electrons in the metal).
Finally in one special case you want a high cross-section - at very high frequencies the 'skin effect' means that current only really goes through close to the surface so you need wires with high surface area (but generally they are hollow).
You might also want to have a look at this question about how electrons behave in a metal when a potential difference (or voltage) is applied.