While argument is canonically undefined at the origin, since $u(r,\theta) = u(0)$ we could define $u(0, \theta):= u(0)$ [i.e., a constant function with respect to $\theta$].
So is $\frac{\partial^2 u}{\partial\theta^2}$ is undefined at the origin (because argument is undefined there) or defined to be zero (because we can define $u(0,\theta)$ as constant function in $\theta$?
If the latter, then suppose that $u$ is harmonic near $z=0$. Does that mean that $$\frac{\partial^2 u}{\partial r^2} + \frac{1}{r}\frac{\partial u}{\partial r}\mid_{r=0} = 0?$$
And what would that even mean, since $r = 0$ and the equation $x + \infty = 0$ has no solution in $\mathbb{R}\cup\{\infty\}$? Not to mention if $\frac{\delta u}{\delta r} = 0$.
Yet there is no reason to believe that a harmonic function behaves differently at the origin than at any other point. What am I missing here?
Generally polar coordinates are not defined at $r=0$, since this is a singular point in the change of coordinates between polar and Cartesian. You can say something about the limit of $\partial u/\partial \theta$ as $r\to 0$, if $u$ is smooth in Cartesian coordinates. In this case you could write
$$u(r,\theta) = f(r\cos\theta,r\sin\theta)$$
for a function $f$ which is smooth and then for $r\neq 0$ we have
$$\frac{\partial u}{\partial \theta} = r\nabla f(r\cos\theta,r\sin\theta)\cdot (-\sin \theta,\cos\theta).$$
So if $r\to 0$, regardless of $\theta$, we have that $\frac{\partial u}{\partial \theta} \to 0$. The same thing holds for higher derivatives of $u$ in $\theta$. So if your function is smooth in Cartesian coordinates, taking $\frac{\partial^k u}{\partial \theta^k}=0$ at $r=0$ is sensible, since this agrees with the extension by continuity.
For your second question, if the function is harmonic near $z=0$, but not at $z=0$, then you cannot say anything at $z=0$. For example, the function $u(r,\theta) = \log(r)$ is harmonic away from $r=0$ (indeed, this is the fundamental solution of Laplace's equation). It satisfies
$$\frac{\partial^2 u}{\partial r^2} + \frac{1}{r} \frac{\partial u}{\partial r} = 0$$
for $r\neq 0$. But there is no sense in which this holds at $z=0$, since the function and all its derivatives are undefined there.
For an even simpler example, take $u(r,\theta)=r\cos\theta$, which is harmonic everywhere in Cartesian coordinates (it is just $f(x,y)=x$). In this case, Laplace's equation in polar coordinates is
$$\frac{\partial^2 u}{\partial r^2} + \frac{1}{r} \frac{\partial u}{\partial r} + \frac{1}{r^2} \frac{\partial^2 u}{\partial \theta^2} = 0.$$
To make sense of it at $r=0$, we'd have to multiply both sides by $r^2$ to get
$$r^2\frac{\partial^2 u}{\partial r^2} +r\frac{\partial u}{\partial r} + \frac{\partial^2 u}{\partial \theta^2} = 0.$$
Sending $r\to 0$ we get just
$$(*) \ \ \ \frac{\partial^2 u}{\partial \theta^2} = 0.$$
The function $u(r,\theta)=r\cos\theta$ clearly satisfies this (at $r=0$), based on our discussion above, but this is not specific to harmonic functions anymore. Any smooth function in Cartesian coordinates satisfies (*), so the equation loses its meaning at $r=0$.