Is it possible to express various nonlinear motions as straight lines in transformed spacetime? I am trying to understand simple examples of space-time curvature.
Assume for the moment that $c$ is infinite (classical curvature due to Newton's laws). Also, I will only consider 1+1-dimensional space for simplicity: $(x,t)$. 
Case I. I will consider uniform acceleration due to gravity g along the positive x-direction pervading the entire space. Then all particles will have $$x-x_0 = \frac{1}{2}g(t-t_0)^2.$$
For this case, we can use the transformation 
\begin{align}
x' &= x - \frac{1}{2}g(t-t_0)^2 \\
t' &= t\end{align}
to get a transformed space-time $(x',t')$ such that any path of the form 
$$x-x_0 = \frac{1}{2}g(t-t_0)^2$$
in the original co-ordinate system is equivalent to the form 
$$x' = vt' + x_0'$$
representing uniform motion in the new $(x',t')$ coordinate system. Thus, if space-time is warped as described by the transformation from $(x,t)$ to $(x',t')$ system, all objects just follow a straight line in the transformed system.
Case II. Now I will consider a slightly more complex scenario. Here, 
$$g(x) = -\omega^2 x.$$
Can we obtain transformation $x'=x'(x,t)$ and $t'=t'(x,t)$ such that uniform motion in $(x',t')$ is equivalent to Simple Harmonic motion in $(x,t)$?
Case III. Now assume that acceleration due to gravity is $g$ for $-1 \le x \le 1$ and is 0 everywhere else. What about this field?
I don't have any experience in differential geometry or any obscure mathematics. My current goal is to see how much we can understand relativity just with  simple mathematics. 
 A: The Lorentz transformation simplifies to the classical transformation for all your cases since you took $c$ as infinity.  There is no time dilation. The transform is just an identity multiplication for all 3 cases with:
$$x' = x - vt$$ and $$t' = t$$
Now that you are in the classical world in one dimension, your first case is merely a tilted line with objects rolling down the slope, no matter from where on the line you look.  Like standing on a hill watching a ball roll down toward you, or away from you.
In second case: 
$$x' = (-w^2)x' = (-w^2)(x - vt)$$ and $$t' = t$$
In the third case: acceleration due to gravity will be g also for moving observer but just seen at coordinates shifted from $-1 <= x <= 1$ becomes $-1 <= x'+vt' <= 1$ or $-1-vt' <= x' <= 1-vt'$ and $0$ everywhere else
A: If you are allowing non-linear transformations of space-time (as you have done in your Case I), then I don't see why you can't write any motion of an object as x' = 0. i.e. Suppose the motion of the object in the (x,t) coordinates can be written as f(x,t) = 0 for some non-linear function f. Then, by setting x' = f(x,t), t' = t, the motion in (x',t') can be simply written as x' = 0. For example, the answer to your second case can be written as: Transform (x,t) to (x' = A*sin(wt+d),t'=t), then the motion is simply x'=0. 
A: Ok I solved case 2. Let $t'=\tan^{-1}(\omega t)$ and $x'=x\sqrt{1+t'^2}$. Also observe that $x=A\cos \omega t + B\sin \omega t$ to represent simple harmonic motion. We get $x'=A+Bt'$. So only case 3 remains open. 
Also as u can see, I just did guesswork, without proper techniques for how to solve this in general. 
