When a particle of mass $m$ is acted upon by a force $F_1(t)$, its acceleration is given by Newton's second law as $$\frac{dv_1}{dt}=\frac{F_1(t)}{m}.$$ Similarly, when acted upon by a force $F_2(t)$, its acceleration is given by $$\frac{dv_2}{dt}=\frac{F_2(t)}{m}.$$
These two equations have the formal solutions $$v_1(t)=v_1(0)+\frac{1}{m}\int_{0}^{t}F_1(t')dt'$$ and $$v_2(t)=v_2(0)+\frac{1}{m}\int_{0}^{t}F_2(t')dt'$$
When both the forces are present simultaneously, the acceleration is given by $$\frac{dv}{dt}=\frac{F_1(t)+F_2(t)}{m}$$ which has the solution $$v(t)=v(0)+\frac{1}{m}\int_{0}^{t}F_1(t')dt'+\frac{1}{m}\int_{0}^{t}F_2(t')dt'$$
Therefore, if $v(0)=v_1(0)+v_2(0)$, $$v(t)=v_1(t)+v_2(t)$$ i.e., the instantaneous velocity when both the forces are present, is equal to the sum of the instantaneous velocities caused by each force acting separately. Let me call this superposition of solutions.
Now, consider the differential equations with forces that depend both on time and linearly on velocity: $$\frac{dv_1}{dt}=\frac{F_1(v_1,t)}{m}, ~~{\rm and}~~ \frac{dv_2}{dt}=\frac{F_2(v_2,t)}{m},$$ and let the condition $v(0)=v_1(0)+v_2(0)$ is met at $t=0$. In this case, the instantaneous velocity $v(t)$ in presence of both the forces is, in general, not obtained by adding the instantaneous velocities, $v_1(t)$ and $v_2(t)$, caused by $F_1$ and $F_2$ separately. For example, let $F_1(v_1,t)=-m\gamma v_1(t)$ and $F_2(v_2,t)=F_0\cos\omega t$ where $\gamma,\omega$ are real constants. Clearly, $$v_1(t)=v_1(0)e^{-\gamma t}, v_2(t)=v_2(0)+\frac{F_0}{m}\frac{\sin\omega t}{\omega}.$$
But the solution to $$\frac{dv}{dt}=\frac{F_1(v_1,t)+F_2(v_2,t)}{m}= -\gamma v+\frac{F_0}{m}\cos\omega t$$ is not given by $$v(t)\neq v_1(t)+v_2(t)$$ or $$v(t)\neq v_1(0)e^{-\gamma t}+\frac{F_0}{m}\frac{\sin\omega t}{\omega}+v_2(0).$$
Why in the second case, the superposition of solutions fails? How can we see/argue more clearly that the solution $v(t)$ will not be $v(t)=v_1(t)+v_2(t)$?
There is absolutely nothing wrong with changing variables $$ v\equiv v_1+v_2 , ~~~~ u= v_1-v_2, $$ and simplifying your equations with them, if you can.
But you simply did it incorrectly: $$\frac{dv}{dt}=\frac{F_1(v_1,t)+F_2(v_2,t)}{m}= -{\gamma\over 2}( v +u)+\frac{F_0}{m}\cos\omega t \\ \neq -\gamma v+\frac{F_0}{m}\cos\omega t, \\ \frac{du}{dt}= -{\gamma\over 2}( v +u)-\frac{F_0}{m}\cos\omega t, $$ so the two linear equations do not decouple.
It is no surprise, then, that the correct solution you got does not solve a wrong equation; it solves the correct linear equations written.
Nevertheless, there is an obvious way to decouple them. You might, instead, define $z\equiv \ln v_1$, so that $$ \frac{dz}{dt}=-m\gamma, $$
which is of the form of your first example, where the variables do decouple, in which case, $z+v_2$ is a fine solution as in your correct starting example.
Takeaway: linearity works just fine, but you must utilize it correctly.