1. If the outcomes are very similar, then you can turn off large deflection and spend less time waiting for the solution. The classic example is a 1m long horizontal cantilevered beam with an end displacement 1 m downward. The linear solution is the tip follows a straight line 1 m down, making the beam lenth over 1.4 m long! The nonlinear solution (large deflection on) is the beam doesn't get any longer, the tip follows a curve on its way down. These two solutions are obviously different.
2. It's fine to have roughly modeled components that are there to provide loads and supports to the components of interest and to ignore the stress results on those components because they are not modeled correctly. The way to ignore the stress on those components is to exclude those bodies when creating a stress result. There is a Scope line that defaults to All Bodies. Select All with a Ctrl-A then with the Ctrl key held, click the body you want to remove, leaving only the bodies of interest and you will only get stress results plotted on those bodies.
However, at some point, it would be worth creating a correct model of the bike frame to check that the new loads on the bike frame are not excessive.
3. If your model solves with weak springs off, then turn them off and you can't look at them. Only if your model will not solve without their help do you have to evaluate their contribution to the solution. I almost never need weak springs. But if you need them, then a 2% contribution would be acceptable.
4. It means there is no stress stiffening in this model. The classic example of stress stiffening is a guitar string. If you have a model of a long thin beam fixed at both ends, it will have an extremely low first natural frequency. If you apply a large amount of tension in a Static Structural model, then link that solution to the Modal setup cell, now you can get high frequencies in the audible range.
5. Yes, but the results are only as good as the inputs, as you mention in point #6.
6. The static model can be a fairly close representation of the peak stress in the frame if you have the right constraints, forces, torques, masses and accelerations applied.
You had an acceleration load to represent the braking and turning acceleration, but I wonder why the braking and cornering accelerations are so low. Dividing the number by 9.8 m/s^2 converts it into a G-force scale. You have only 0.02 and 0.01 G of acceleration. That is tiny. Is this to simulate taking your grandmother out for a Sunday drive? If this bike was in a race, it could easily reach braking and cornering accelerations of 0.3 G. I created an acceleration load shown below.
Here is another idea, when turning left while braking, most of the weight would be on the front and outside wheels, while the rear wheel would carry little weight. Therefore, I recommend that the outside wheel take the YZ = 0 constraint, while the rear wheel takes the Z=0 constraint. This lets the front and outside wheel support the sideways cornering force, which puts more stress into the sidecar frame.
The distance of the side-car tire contact point from its axle causes a bending moment into the side-car frame due to cornering accelerations. You should use a remote displacement to the tire contact point instead of at the axle for rueda-webo_YZ=0. Without that offset of the remote displacements, this model is less severe than it should be. I moved the remote point down by 300 mm on the front and rear wheels, and down by 200 mm on the sidecar wheel.
7. Safety factors are used to cover the unknowns in the inputs, the materials, the geometry, the connections, etc. I don't know what to use for that. It depends on the severity of the consequence of failure. Follow the practices in the literature.