Let’s choose a cylindrical coordinate system $\rho,\theta,z$ along the axis of the wire. We’ll assume that the wire is the the positive conductor in some electric circuit, and therefore has a small nonzero net charge; see e.g. the illustration at this other answer.
Model the electric field as
\begin{align}
\vec E_\text{inside} &= \frac1\sigma \vec J = E_\text{in}\hat z
\\
\vec E_\text{outside} &= E_\text{out} \frac R\rho \hat \rho
\end{align}
where $\vec J$ is the current density,
$\sigma$ is the conductivity of the wire,
$R$ is the wire’s radius, and the field magnitudes $E_\text{in}$ and $E_\text{out}$ are the solutions to freshman homework problems.
Actually, we can see that this approximation is not strictly allowed. A surface charge, such as on the edge of our wire, introduces a discontinuity in the field normal the surface, the $\hat \rho$ direction. However the field components parallel to the surface, in the $\hat\theta$ and $\hat z$ directions, must be continuous. So our outside field must actually be
$$
\vec E_\text{outside} = E_\text{out}\frac R\rho \hat\rho + E_\text{in} \hat z
$$
Our assumption that the field outside the wire is radial is therefore equivalent to an assumption that the conductivity $\sigma$ of the wire is large.
The magnetic field at some distance $\rho$ from the axis is
$$
\vec B = \frac{\mu_0}{2\pi} \frac{I_\text{enclosed}}{\rho} \hat \theta
$$
Outside of the wire, $\rho > R$, the enclosed current is just the entire current, $I = \pi R^2 J$. Inside the wire, only the inner part of the current contributes, and the field becomes
$$
\vec B_\text{inside} = \frac{\mu_0}{2\pi}\frac{\pi\rho^2 J}{\rho}\hat\theta \propto \rho\hat\theta
$$
We can find the direction of the Poynting vector $\vec S \propto \vec E \times \vec B$ everywhere. Outside of the wire there is a component proportional to
$$\vec S_\text{downstream} \propto E_\text{out} \hat\rho \times \hat\theta = E_\text{out} \hat z$$
That is to say, the power transport outside of the wire is parallel to the direction of the current in the “supply” wire of the circuit. The current-return wire will have a small net negative charge, $\vec E{}_\text{outside}^\text{return} \propto -\hat\rho$, and energy will flow antiparallel to the current — also towards the load.
There is also a radial power flow proportional to
$$
\vec S_\text{radial} \propto E_\text{in} |\vec B| (-\hat\rho)
$$
which is responsible for heating the wire. To find how much energy is deposited per unit volume, let’s consider a cylindrical shell with radius $r$, thickness $d\rho$, length $L$, and volume $2\pi rL\,d\rho$.
The power deposited in this shell is given the difference between the power entering from its outside and the power exiting towards its inside:
\begin{align}
P &=
\int_\text{outer} \vec S \cdot d\vec A
-
\int_\text{inner} \vec S \cdot d\vec A
\\
&= S_\text{outer} A_\text{outer} - S_\text{inner} A_\text{inner}
\end{align}
Outside of the wire, we have $S \propto B \propto r^{-1}$, and area $A \propto r$, so no energy is deposited in the current-free space outside of the wire. However, inside the wire we have
$$
\vec S \cdot d\vec A \propto BA \propto r^2
$$
The power in a shell with thickness $d\rho$ therefore goes like
\begin{align}
B_\text{outer} A_\text{outer}
-
B_\text{inner} A_\text{inner}
& \propto
(r + d\rho)^2 - r^2 \approx 2r\, d\rho
\end{align}
Since the power per unit volume and the volume itself are both proportional to $r\,d\rho$, we have uniform power deposition in the metal.
You ask why the wire is hotter at the center than at the edges. This is a heat-transport problem: the heat can only leave the wire through its surface, so the surface must cool before the interior.
