The angular momentum along a certain axis is not the full momentum, which has $3$ components. Thus, $\hat L_z$ along $\boldsymbol{\hat z}$ is not the "full angular momentum" which would also include $\hat L_y$ and $\hat L_x$.
Suppose for instance the electron has a definite value of $\hat L_z$, i.e. the state $\psi(r,\theta,\phi)$ is such that
$$
\hat L_z\psi(r,\theta,\phi)=\hbar m \psi(r,\theta,\phi)\, .
$$
Then $\langle L_z\rangle= \hbar m$ and $\Delta L_z=0$: the last two equalities validate the statement that the electron has definite value of $\hat L_z$ as there is no fluctuation of that value from the average value $m\hbar$.
The uncertainty relation for angular momentum then states that
$$
\Delta L_x\Delta L_y\ge \frac{1}{2}\vert \langle L_z\rangle\vert =\frac{1}{2}\vert m\vert \hbar
\tag{1}
$$
so that neither $\Delta L_x$ nor $\Delta L_y$ can be $0$, and consequently that the electron in the state $\psi(r,\theta,\phi)$ does not have a well-defined value of $\hat L_x$ or $\hat L_y$.
If the state $\psi(r,\theta,\phi)$ has well defined angular momentum in the sense that
$$
\hat L^2\psi(r,\theta,\phi)=\ell(\ell+1)\hbar^2 \psi(r,\theta,\phi)\, ,
$$
then the magnitude squared of $\vec L$ is well defined, and its $\boldsymbol{\hat z}$-component is well-defined, leaving the tip of $\hat yL_y+ \hat x L_x$ to lie somewhere on a circle in the $xy$ plane at height $m\hbar$, as illustrated below for $\ell=2$.
The tip of the angular momentum must be found equally likely anywhere on the circle. To see this,
you can also use the cyclicity of (1) to find
\begin{align}
\Delta L_x\Delta L_z&\ge \frac{1}{2}\vert \langle L_y\rangle\vert =0\, \\
\Delta L_y\Delta L_z&\ge \frac{1}{2}\vert \langle L_x\rangle\vert =0\,
\end{align}
since $\Delta L_z=0$ by assumption on the state $\psi(r,\theta,\phi)$,
showing that $\langle L_x\rangle = \langle L_y\rangle =0$ and justifying the random location of the tip on the circle.
