What are the forces present in a coordinated turn?
To get a deeper understanding beyond the superficial, it helps to break down the forces and moments present on an aircraft that may affect its rigid-body motion:

Aerodynamic forces: These are the forces and moments exerted by the airflow on the aircraft

Ground forces: These are the forces and moments exerted by the ground on the aircraft, transmitted through tyres and landing gear. Not applicable when it's flying.

Propulsion: Forces and moments due to direct thrust. For simplification, let's assume that thrust acts inline with the forward axis.

Gravitational forces: Gravity pulls the aircraft towards the ground. It's rather special because everything, be the aircraft structures, you, me, or the accelerometers, get pulled at the same rate ($g$)1. This is distinctly different than the other two types of forces, which are only present when airflow affects exposed areas, or when contact with the ground has been made.

Inertial forces: These are fictitious forces and moments that are required to maintain non-uniform motion. This includes your centripetal force. Inertial forces must always be equal to the sum of all the aforementioned external forces.


First of all, let's consider any turn a steady-state maneuver (thereby ignoring the transients like rolling in and rolling out), which means that the vector sum of all the external forces, including gravity, must sum to the inertial forces. As you've correctly pointed out, the sum of aerodynamic forces + gravitational forces must be equal to the centripetal force, which in the turning plane is provided by lift, side force and gravity. This must hold true for any steady-state turn, whether coordinated or not.

There are two ways to define a coordinated turn. With all engines operating, they are approximately equivalent:

A zero-sideslip turn
A ball-centered turn: we are going to use this definition

What ball-centered means is that there are no aerodynamic forces acting laterally on the airplane: lift must provide all the centripetal force required. Ball centered provides the best average feel for the occupants, since the forces are directly inline with the floor, and there's no side force causing sway. Since everything feels gravity at the same rate, occupants or the ball cannot detect gravity.
For illustration:


For the more math oriented, in inertial frame, Newton's second law is stated as:
$$\vec{F_{i}}+m\vec{g_i}=m\frac{d\vec{V_i}}{dt} \tag{1}$$
The entire right hand side is considered to be inertial forces. However, if the measurements occur in a rotating frame (on an airplane, for example), then we need to express everything into the body frame. The left-hand side is easy:
$$\vec{F_{b}} = C_{bi}\vec{F_{i}} \tag{2}$$
$$\vec{g_b} = C_{bi}\vec{g_i} \tag{3}$$
where $C_{bi}$ is the rotation matrix transforming a vector from inertial frame to body frame.
The right-hand side requires some adjustments, because the Euler angles of the body themselves are changing:
$$\frac{d\vec{V_b}}{dt}=\frac{d(C_{bi}\vec{V_i})}{dt}$$
Apply chain rule, and we have:
$$\frac{d\vec{V_b}}{dt}=\frac{dC_{bi}}{dt}\vec{V_i}+C_{bi}\frac{d\vec{V_i}}{dt} \tag{4}$$
It can be mathematically shown the following identity:
$$\frac{dC_{bi}}{dt}\vec{V_i} = -\vec{\omega_b} \times \vec{V_b} \tag{5}$$
Substitute (5) into (4), and we get:
$$C_{bi}\frac{d\vec{V_i}}{dt} = \frac{d\vec{V_b}}{dt} + \vec{\omega_b} \times \vec{V_b}$$
Finally, if we multiply both sides of (1) by $C_{bi}$ and simplify with (2), (3) and (5), we get:
$$\vec{F_{b}}+m\vec{g_b}=m \left( \frac{d\vec{V_b}}{dt}+\vec{\omega_b} \times \vec{V_b} \right) \tag{6}$$
(6) is the Newton's second law in a rotating frame. The entire right-hand side are still inertial forces (same as (1)), except now we also have the cross-product, which produces the centripetal inertial forces:
$$\vec{\omega_b} \times \vec{V_b} = \begin{bmatrix}p \\ q \\ r\end{bmatrix} \times \begin{bmatrix}u \\ v \\ w\end{bmatrix} = \begin{bmatrix} qw - rv \\ ru - pw \\ pv - qu \end{bmatrix} \tag{7}$$
You will readily identify terms like $ru$, which is very close to the familiar $a_c=\omega V$ for a restricted 2D centripetal motion (whence $p=0$ and $q=0$).
One final note, the right-hand side is fictitious in that they are not real forces! They are a result of the kinematic motion itself, and must require the left-hand side (which are the real forces), to sustain.

1: Technically, this is only true locally, because Earth is a sphere and does not exert a uniform field at different altitudes. But at the range of altitudes that airplanes will be flying, this is a rather good approximation.