How to deduce the dynamic general equations for a non-symmetric level turn manoeuvre?
Your equations seem to be in wind-axis, and look a little weird from my literature (referencing Stevens and Lewis, Aircraft Control and Simulation). I'm more used to specifying mechanics from the body-axis.

Legend: $\theta$ is Euler pitch angle, $\phi$ is bank angle, $\psi$ is yaw angle, $p$ is body roll rate, $q$ is body pitch rate, $r$ is body yaw rate, $\alpha$ is angle of attack, $Y$ is aerodynamic side force (in stability-axis), $u$ is body forward speed, $v$ is body lateral speed, $w$ is body vertical speed.

A. Thrust aligned with body-axis

I will assume:


Thrust is aligned with the body-axis.
All aerodynamic forces are aligned with the stability-axis (see this question for the rationale).
No wind, so that the velocity in body-axis is the relative air velocity ($\alpha=\tan^{-1}(w/u)$,$\beta=\sin^{-1}(v/\sqrt{u^2+v^2+w^2})=\sin^{-1}(v/V)$).


For the general 6-dof, we have:

$$T-D\cos\alpha+L\sin\alpha-mg\sin\theta=m(\dot{u}+qw-rv)$$
$$Y+mg\cos\theta\sin\phi=m(\dot{v}+ru-pw)$$
$$D\sin\alpha-L\cos\alpha+mg\cos\theta\cos\phi=m(\dot{w}+pv-qu)$$

In a coordinated turn, the ball is centered, which means that the gravity component cancels out the centripetal force, and the aerodynamic forces in the y-axis must be zero. 

Furthermore, in the inertial frame, we should only see a non-zero yaw rate ($\dot{\psi}$) for a steady turn. So translating this back into body-axis, we have:

$$p=-\dot{\psi}\sin\theta$$
$$q=\dot{\psi}\cos\theta\sin\phi$$
$$r=\dot{\psi}\cos\theta\cos\phi$$

To make the matter somewhat clearer, we can choose to make small angle assumptions on $\alpha$ and $\theta$, and $u\approx V$, and eliminate all second order terms. This brings us to:

$$T-D+L\alpha-mg\theta=m(\alpha V\dot{\psi}\sin\phi- \beta V\dot{\psi}\cos\phi)$$
$$mg\sin\phi=mV\dot{\psi}\cos\phi$$
$$-L+mg\cos\phi=-mV\dot{\psi}\sin\phi$$

From geometry, a particle tracing out a constant circular path with a rate of $\dot{\psi}$ will have a radius (i.e. turn radius) of: $R=V/\dot{\psi}$. So the second equation is now:

$$tan\phi=\frac{V\dot{\psi}}{g}=\frac{V^2}{gR}$$

We can define a stability-axis load factor ($n_{z_s}=\frac{L}{mg}$), which would closely approximate the body load factor for small AOA. The third equation can be expressed as:

$$n_{z_s}=\cos\phi+\frac{V\dot{\psi}}{g}\sin\phi=\sec\phi$$

At this point, we've completely reproduced the performance equations relating load factor, bank angle and turn radius from the 6-dof equations.

B. Thrust asymmetric with body axis

Let's say that the thrust vector is angled with the body axis by a pitching angle of $\epsilon$ and a yaw angle of $\nu$.

So in the body-axis, the thrust vector is:

$$\boldsymbol{T_b}=\begin{bmatrix} \cos\epsilon\cos\nu &amp; -\cos\epsilon\sin\nu &amp; \sin\epsilon \\ 
\sin\nu &amp; \cos\nu &amp; 0 \\
-\sin\epsilon\cos\nu &amp; \sin\epsilon\sin\nu &amp; \cos\epsilon \end{bmatrix}
\begin{bmatrix}T \\ 0 \\ 0 \end{bmatrix}$$

We now have modified 6-dof:

$$T\cos\epsilon\cos\nu-D\cos\alpha+L\sin\alpha-mg\sin\theta=m(\dot{u}+qw-rv)$$
$$T\sin\nu + Y+mg\cos\theta\sin\phi=m(\dot{v}+ru-pw)$$
$$-T\sin\epsilon\cos\nu + D\sin\alpha-L\cos\alpha+mg\cos\theta\cos\phi=m(\dot{w}+pv-qu)$$

Again assume that the ball is centered, thus $T\sin\nu = -Y$. The conclusion from the second equation (relating bank angle to yaw rate and turn radius) would be the same as before.

The third equation can be reduced to:

$$\frac{1}{mg}(T\sin\epsilon\cos\nu+L)=\sec\phi$$