I'll assume that you have a positive susceptibility, ie the induced polarisation is in the same direction as the applied electric field (and not along its opposite direction). Mathematically, assuming an isotropic, linear dielectric, you have:
$$
\vec P = \epsilon_0\chi\vec E
$$
with $\chi>0$, which gives $\epsilon_r = 1+\chi>1$ hence an attenuation of the Coulomb force.
The keyword is electric "shielding," which is responsible for this attenuation. It is true that the electric dipole field is roughly parallel to dipole moment around the origin, which is why you could expect an enhancement of the field. However, if you are careful, this enhancement will be valid only outside the dielectric, while inside it will rather be attenuated.
One way to see it is that the electric field $\vec E$ at position $\vec r$ generated by an ideal dipole of moment $\vec p$:
$$
\vec E = \frac{1}{4\pi\epsilon_0 r^5}(3\vec r\cdot \vec p)\vec r - r^2\vec p))-\frac{1}{3\epsilon_0}\delta^{(3)}(\vec r)\vec p
$$
with the last term often omitted because the formula is usually applied at large distances $r\to \infty$. However, inside dielectrics, it is very important and the contribution of continuum of dipoles of the dielectric gives an additional non-negligible contribution. Note the sign which is the cause of the shielding, ie the attenuation.
Intuitively, this effect is already hinted when considering two separated oppositely charged points. Between the points, the filed line goes from the positive charge to the negative charge, so the field there is roughly in the opposite direction with respect to the dipole moment.
Another example would be a uniformly polarised ball with polarisation $\vec P$. You can view it as two slightly shifted uniformly, oppositely charged balls. Since you know the field inside a uniformly charged ball of charge density $\rho$ is $\vec E = \frac{\rho}{3\epsilon_0}\vec r$, you get a uniform field inside the polarised ball:
$$
\vec E = -\frac{\vec P}{3\epsilon_0}
$$
Btw, this is how you retrieve the delta term for the first equation. The sign can be figured out intuitively as well. Knowing that the fields line are straight (which only a detailed calculation can give you), since they must go from the positive charges to the negative ones, the field must be opposite to polarisation (which goes from negative to positive).
Btw, this apparent generalised Coulomb force for dielectrics is to be used carefully as many assumptions are involved (homogenous, isotropic, linear dielectric). More importantly, it assumes that the dielectric encompasses all of space, or rather more realistically that edge effects are negligible. If any of these assumptions fail, you'll need to solve the Maxwell equation.
Hope this helps.
