Why does water falling slowly from a tap bend inwards?

You can actually predict the shape of the profile precisely using the arguments you mention above, which are by and large correct. To do so, you can make the following assumptions:

• Neglect viscosity (not a great assumption, but it's a start).
• The pressure is the same everywhere in the fluid—the edges are free surfaces, so this is reasonable.
• The flow is axially symmetric (i.e. the top-down cross section is always circular).

If you do this, and take the location of the faucet as the origin, you can then state the relationship between the gravitational potential energy and the flow speed using Bernoulli's equation as:

$$\rho g h + \rho \frac{1}{2}v^2 = \rho \frac{1}{2}v_0^2$$

where $v$ is the speed of the fluid as a function of height $h$, $\rho$ is density, and $v_0$ is the speed at which the water leaves the faucet.

Solving for $v$, you'll find that:

$$v = \sqrt{v_0^2 - 2gh}$$

As the fluid moves further down (i.e. as $h$ becomes further negative), the speed increases as you'd expect.

Then you can use conservation of mass for the rest. Assuming steady flow, you'll find that

$$A_1 v_1 = A_2 v_2$$

for any two cross-sections of the flow. Using the cross-sections at the faucet and another arbitrary cross-section, and declaring the faucet radius as $r_0$, you'll find:

$$\pi r_0^2 v_0 = \pi r^2 v$$ $$\pi r_0^2 v_0 = \pi r^2 \sqrt{v_0^2 - 2gh}$$

Solving for the radius $r$, you find up getting the following expression:

$$\boxed{r(h) = \frac{r_0 \sqrt{v_0}}{(v_0^2 - 2 g h)^{1/4}}}$$

This drop in the radius as the height decreases is consistent with your illustrations. For example, here is what I analytically determine as the flow profile when I use standard values for a bathroom sink faucet flow ($r_0 = 1.5$ centimeters, $v_0 = 0.134$ meters per second, and $g = 9.81$ meters per second squared):

Notice that the flow profile becomes effectively straight at distances observable in your common bathroom sink (4 inches or so). This is consistent with your observations.

After a certain point, the stream becomes so thin that surface tension effects along with shearing at the air-water interface begin to destabilize the shape and cause it to break up into droplets. In addition, the flow becomes turbulent after a certain distance from the faucet, so this prediction is only accurate for the early stages of such a flow (i.e. for "small" $h$).

To enlarge slightly upon @aghostinthefigures' excellent exposition, for small gravity-driven jets the flow does not go turbulent- instead, it is subject to rayleigh instability when its cross-section gets small enough for surface tension forces to become dominant. At that point, any small perturbation of the jet will cause it to spontaneously break up into individual droplets before the flow in the jet has the opportunity to become turbulent.