The problem with your a approach is, that this is a statically indeterminate system. Basically, you could for example turn one fixed bearing into a simple support and the structure would still be statically stable.
Remove another reaction force and the system turns into a mechanism.
Thus the degree of static indeterminacy is $n=1$.
As a consequence, the reaction forces depend on the materials elastic properties and cannot be determined simply by solving the forces and moment equations.
My approach was using the force method, assuming the elastic modulus $E$ and the area moment of inertia $I$ to be constant throughout the entire beam.
For this more general case of beam-lengths $a$ and $b$ I got the following solutions:
$$ A=F\frac{b^3}{a^3+b^3} \qquad B=F\frac{a^3}{a^3+b^3} $$
Thus for the case $a=L$ and $b=2L$:
$$ A=\frac{8}{9}F \qquad B=\frac{1}{9}F $$
PS: It may look like the solution is not depending on the elastic properties of the beam, but this is due to the assumption of a homogeneous beam, allowing me to cancel out $EI$ during my calculations.
force-method: (see attachment)
The principle behind the force method is, that you remove e.g. the moment restraint at point $B$ (replace it with a simple support, so it becomes determinate) and then calculate the moment distribution of this determinate system.
Then, by implementing a unitary moment $X_1=1$, you calculate the required moment $X_1=M_B$ to satisfy the conditions of a fixed restraint: no rotation at $B$ $\quad \to\delta_1=\delta_{10}+X_1\cdot\delta_{11}=0 $
note: $\delta_1$ is the rotation at $B$, which must be zero, $\delta_{10}$ is the rotation caused by $F$ in the reduced system (marked as GS), $X_1\cdot\delta_{11}$ is the rotation caused by the moment $X_1$ (marked as ÜS), thus when you solve for $X_1$ you get $M_B$
