Here is a first attempt:
Assume that $g:\quad (x,y)\mapsto g(x,y):=\sum_{j,k\geq 0} a_{jk}\, x^j y^k$ is a real analytic function defined in the unit disk $D:=\{(x,y)\ |\ x^2+y^2<1\}$ and that $f(r,\phi):=g(r\cos\phi, r\sin\phi)\qquad(r\geq0,\ \phi\in{\mathbb R})\ .$ What can we say about $f\,$?
Right away we can say that $f$ is a real analytic function of $r$ and $\phi$ defined for $-1 and all $\phi\in{\mathbb R}$; furthermore $f$ is $2\pi$-periodic in $\phi$, and $f(-r,\phi)\equiv f(r,\phi+\pi)$. But there is more to it.
Putting $z:=x+i y=r e^{i\phi}\ ,\quad \bar z:=x-i y= r e^{-i\phi}$ we can write $g$ in the form $g^*(z,\bar z)=\sum_{j,k} c_{jk}\, z^j \bar z^k=\sum_{l\geq 0} r^l \ T_l(\phi)$ where $T_l(\phi)=\sum_{j,k\geq 0;\ j+k=l} c_{jk}e^{(j-k)\phi}\ .$ Therefore we can conclude the following: The polar representation $f$ of $g$ is necessarily of the form $f(r,\phi)=\sum_{l=0}^\infty r^l\ T_l(\phi)\ ,$ where $T_l(\cdot)$ is a trigonometric polynomial of degree $\leq l$ that satisfies $T(\phi+\pi)=(-1)^lT(\phi)$. By going backwards it is easy to see that any such $f$ is the polar representation of a real analytic $g:\ (x,y)\mapsto g(x,y)$.
The $C^m$-case is not so easy. If we want $g$ to be in $C^m$ then $f=g\circ{\rm rect}\ $ will have to be in $C^m$ also. As $g$ is at at least $m$ times differentiable at $(0,0)$ one has $g(x,y)=\sum_{j,k=0}^m a_{jk}\, x^j y^k + o(r^m)\qquad (r\to 0)\ .$ It follows that necessarily $f(r,\phi)=\sum_{l=0}^m r^l T_l(\phi)+o(r^m)\qquad (r\to 0)\ ,$ where the $T_l$ are as before. But this is not enough to guarantee the continuity of the derivatives of $g$. In the case $m=1$ the complete answer looks as follows:
The function $f$ is the polar representation of a $C^1$ function $g$ defined in a neighborhood of the origin iff there are constants $a$, $b$, $c$ and a $C^1$-function $R$ such that $f(r,\phi)=c + a\, r \cos\phi+b\, r\sin\phi +R(r,\phi)\ ,$ $R(0,\phi)=0,\quad R_r(0,\phi)=0, \qquad R_\phi(r,\phi)=o(r)\quad(r\to0).$
Proof. Let $g(0,0)=:c$, $g_x(0,0)=:a$, $g_y(0,0)=:b$, and put $h(x,y):=g(x,y)- ax - by\ , \quad R(r,\phi):=h(r\cos\phi,r\sin\phi)\ .$ The facts $h(0,0)=h_x(0,0)=h_y(0,0)=0$ together with the equations $R_r=h_x\cos\phi + h_y\sin \phi\,\qquad R_\phi=r\ (-h_x\sin\phi + h_y\cos\phi)$ show that the stated conditions on ($a$, $b$, $c$ and) $R$ are necessary, and the dual equations $h_x= R_r\cos\phi-{1\over r}R_\phi\sin\phi\ ,\qquad h_y=R_r\sin\phi +{1\over r}R_\phi\cos\phi$ show that these conditions are also sufficient to guarantee the continuity of $h_x$, $h_y$, resp., $g_x$, $g_y$.