Affine Non Autonomous State Space system

You have a nonzero operating point. If you had chosen an equilibrium operating point, this situation would not have occurred.

With the equation you have given, if you choose the states as \$\left\{x_1=\frac{g m}{K_d}+V_z,x_2=v_1^2,x_3=v_2^2,x_4=v_3^2,x_5=v_4^2\right\}\$ you get a linear state equation:

\$ \dot{x}_1=\frac{K \left(x_2+x_3+x_4+x_5\right) C_m}{m}-\frac{x_1 K_d}{m} \$

I am guessing this would upset some other state equation. Then you have to consider all the states and equations together. But, as I said before the best approach is to do this at the linearization step.

You misunderstood state-space representation. While the standard form you've posted indeed targets linear systems but this doesn't mean the representation can't handle nonlinear system. Probably one of powerful tools of this representation is its capability for handling nonlinear system. Let's see how the system is represented in the state space (i.e. no matrix form).

Let's define \$q_1=x, q_2=\dot{x},q_3=y, q_4=\dot{y},q_5=\theta, q_6=\dot{\theta}\$, therefore, we obtain

$$ \begin{align} \dot{q}_1 &= q_2\\ \dot{q}_2 &= F_E\cos(\psi)\sin(q_5) + F_E\sin(\psi)\cos(q_5) +F_s\cos(q_5)\\ \dot{q}_3 &= q_4\\ \dot{q}_4 &= F_E\cos(\psi)\cos(q_5) - F_E\sin(\psi)\sin(q_5) - F_s\sin(q_5) \\ \dot{q}_5 &= q_6\\ \dot{q}_6 &= F_E\sin(\psi)(l_1+l_n) - l_2F_s\\ \end{align} $$ Now the system can be represented generally as

$$ \dot{q} = f(q, u), \quad q \in \mathbb{R}^3, u \in \mathbb{R}^3 $$ which can be solved numerically. I can solve the system if you provide me with the actual parameters of the system.