Understanding stability of numerical ODE methods from just linear problem So in general for ODE we come up with a numerical method for
$y'=f(t,y)$
Then apply the numerical method to $y'=\lambda y$, where $\lambda$ is some scalar
My question is how can we claim the stability criteria we get for a numerical method from solving $y'=\lambda y$ have any influence in the general case $y'=f(t,y)$
 A: I wouldn't say that it always does. But, if a numerical method can't even handle a linear problem, then it's probably going to have trouble on a more significant problem. Generally, the different types of stability (zero stability, absolute stability, etc.) are more so a means of crossing methods off the list, so to speak, or to show what context the method is viable. Being absolutely stable, for a given range of $\lambda$, doesn't necessarily say whether a method is good, but more so if it's not (or, to be less cavalier, how restrictive it is). 
Take backward Euler. Its stability region is pretty large (it contains the left half-plane; such a method is called $A$-stable), so it's a solid option for the linear problem. Take the explicit midpoint/leapfrog rule. Its region is extremely restrictive (open interval from $-i$ to $i$), so we often won't consider it for even the linear problem, which does not inspire much hope for a more complicated $f$ (it does have applications when $\lambda$ is purely imaginary, which happens for hyperbolic PDEs, actually, but the general point still stands). 
If the solution of a non-linear problem $y'=f(y)$ is not changing too much relative to the time step (such as when we have a fixed point), then we can do a linear approximation over a small time interval. In this case, our linear stability results can prove useful (note that it depends on how our method deals with the linear system given by evaluating the Jacobian at the fixed point).
