The book is wrong in that the statement is not necessarily true. Nevertheless, it is right in that there can be an asymmetry.
The short answer is that the thermoelectric effect is responsible for the temperature distribution asymmetry.
The long answer is the following:
In a homogeneous and isotropic current-carrying wire at steady-state, the temperature has to obey the heat equation (1D to keep things simple and ignoring radiation and convection which can be partly circumvented experimentally anyway.) $\kappa \frac{d^2 T}{dx^2}+\rho J^2 -\mu J\frac{dT}{dx}= 0$. I am not yet specifying the boundary conditions.
The first term arises from conservation of energy + Fourier's law. It's essentially the heat conduction. $\kappa$ is the thermal conductivity.
The second term is the Joule heat. $\rho$ is the resistivity and $J$ is the current density. If you were to stop here, the equation would have symmetric solutions unless you assume that, for example, either $\kappa$ or $\rho$ have a non linear temperature dependence in the temperature range the experiment is done. One way to see that, is that if you change $x$ by $-x$ in the equation, it stays the same. And your boundary conditions are symmetric, so there is no reason why the solution would be asymmetric.
Anyway, the third term is the Thomson heat, caused by the Thomson effect. $\mu$ is the Thomson coefficient, and can have either sign. This term causes an asymmetry, regardless of the temperature dependence of the other parameters (you can just assume them constant). One way to see that is to start from a sketch of $T(x)$ when only the Joule heating (and Fourier conduction) are assumed. That plot is symmetric about the middle of the wire and has a maximum right at the center, as you correctly guessed. It looks similar to a parabola in the center region. However, the Thomson heat ($Q_T = -\mu\vec J \cdot \nabla T)$ changes sign about the center of the wire. This is because the temperature gradient, which in about half the wire points in the direction of the current (or against it), points in the opposite direction in the other half, while $\vec J$ maintains its direction. Thus, everywhere in the volume of the wire, there will be a cooling in about half the wire due to the Thomson heat, and a heating in the other half (this is in addition of the Joule heat that occurs through the whole volume too, as the equation describes). But if you focus in a very tiny region near the middle (the temperature gradient is very small, and so is the Thomson heat), you will see that adding heat (i.e. rising the temperature slightly) on a side (of the parabolic $T(x)$) and reducing it on the other side, actually shifts the position of maximum temperature. This is what the Thomson heat does.
The problem with the book is that it assumes that $\mu$ has a particular sign and that it must be so. This is wrong, in reality $\mu$ can have either sign, and if you consider a wire whose material has an opposite Thomson coefficient sign, the maximum temperature will be reached between the middle and the left end of the middle region.
Extra information. It turns out that $\mu=T\frac{\partial S}{\partial T}$ where $S$ is the Seebeck coefficient and whose sign is (somewhat) an indicator of whether a material has a majority of p (holes) or n (electrons) charge carriers. So even in the case where we're sure that electrons are the ones moving, $\mu$ can have either sign, and so the maximum temperature can be reached at either side of the center. This goes against @sammy gerbil's answer. The basis of his answer is correct in that good electrical conductors tend to be good thermal conductors, but this only means that $\kappa$ will have a non negligible, or even dominant, contribution from the electrons as opposed to the phonons. This does not imply that heat flows in the direction of the current, this part is wrong. Heat's direction is determined by the direction of $-\kappa \nabla T + ST \vec J$ where S is the Seebeck coefficient and can have either sign (or even zero).
