$\mathbb{D} = \{z \in \mathbb{C} : |z| < 1\}$. $J_1 = \{e^{i\theta}: \theta \in (0, \pi/2)\}, J_2 = \{e^{i\theta}: \theta \in (\pi/2, \pi)\}, J_3 = \{e^{i\theta}: \theta \in (\pi, 2\pi)\}$
It's not so hard to find a harmonic function $u$ on $\mathbb{D}$ with $u|_{J_1} = 1, u|_{J_2} = 2$, and $u|_{J_3} = 4$
To find such a function, we can transport the problem into the right half plane and take $\tilde u (x,y) = \arctan(y/x)$. Scaling appropriately and adding a constant, we can then use one of many mobius transformations to put it back onto the circle, to get something which is $K$ on an arc of the circle and $0$ everywhere else. Doing it 3 times, we find our function $u$.
But what are the loosest conditions I need so that the solution is unique?
It is necessary that the function be bounded inside the unit disk. But I don't think it's sufficient because we still get many solutions based on our choice of mobius transforms.
One idea is to use maximum principle, but I am not able to make sense of it when the boundary is not continuous.
In general, there are many solutions.
Let $u$ be harmonic and bounded on $\mathbb{D}$ with $u|_{J_1} = 1, u|_{J_2} = 2$, and $u|_{J_3} = 4$. Then $v(z)=u(z)+\rm{Re}\frac{1+z }{1-z}$ is harmonic and unbounded on $\mathbb{D}$ with $v|_{J_1} = 1, v|_{J_2} = 2$, and $v|_{J_3} = 4$. Note that $\lim_{x \to 1} v(x)=\infty$.
On the contrary, if $u$ and $v$ are both bounded solutions, then $u=v$ on $\mathbb{D}$.
Proof. Since $u$ and $v$ are bounded, there is a constant $M>0$ such that $|u-v|<M$. We define $V(z)$ as follows: $$ V(z)=\frac{1}{2\pi}\int_{0 }^{2\pi } P_r(\theta -t)h(e^{it}) dt \quad (z=re^{i\theta }), $$ where $h(e^{it})=M, (t\in [0, \delta ], [\frac{\pi}{2}-\delta , \frac{\pi}{2}+\delta ], [\pi-\delta , \pi+\delta ], [2\pi-\delta , 2\pi])$ and $h(e^{it})=0$ (otherwise), $P_r$ is the Poisson kernel.
Since $u-v-V$ is harmonic and $\limsup_{z \to \partial\mathbb{D} } (u-v-V)\le 0$, we have $u-v-V\le0$ on $\mathbb{D}$ by the maximum principle. Then $u-v\le 0$ because $\lim_{ \delta \to 0} V=0$.
Similarly we have $v-u\le 0$ which concludes $u=v$.