Ways to split nonlinear ODEs in linear part + nonlinearity that may work When one linearizes an ODE around an equilibrium point, say  
$$\begin{pmatrix}x'\\
y'
\end{pmatrix}=\begin{pmatrix}x+y+e^{2y}-1\\
e^{2y}-1
\end{pmatrix}$$
around the point $(0,0)$, then one calculates the jacobian evaluated at the equilibrium, which gives a matrix $A$, and splits this equations in a linear part, give by $A$ and the nonlinear rest (some theorems then tell us, how the behaviot of the linear part governs the behavior of the whole ODE).
Why can't we do this splitting in a different way, e.g., by just collecting "linear terms" ?

Concrete example, to see what I mean:
The jacobian of the above system is
\begin{pmatrix}1 & 1+2e^{2y_{0}}\\
0 & 2e^{2y_{0}}
\end{pmatrix} 
evaluated at $(x_0,y_0)=(0,0)$ being 
\begin{pmatrix}1 & 3\\
0 & 2
\end{pmatrix}
So our original system we can rewrite as $$\begin{pmatrix}x'\\
y'
\end{pmatrix}=\begin{pmatrix}1 & 3\\
0 & 2
\end{pmatrix}\begin{pmatrix}x\\
y
\end{pmatrix}+r\begin{pmatrix}x\\
y
\end{pmatrix}$$
where $r$ is some nonlineare function.
But I could imagine other ways of separating the right-hand side into a linear and a nonlinear part, e.g. in the following way:
$$\begin{pmatrix}x'\\
y'
\end{pmatrix}=\begin{pmatrix}1 & 1\\
0 & 0
\end{pmatrix}\begin{pmatrix}x\\
y
\end{pmatrix}+\tilde{r}\begin{pmatrix}x\\
y
\end{pmatrix}$$
where I just colelcted the linear terms, corresponding to the "$x+y$" term in the original system and used a different nonlinearity $\tilde r$ that swallows all the other terms.
Why is this second way not possible (in the sense that I think the theorems that tell when the linear system governs the whole system, won't work now by applying them to the linear part obtained by this splitting), where lies the intuition behind it being necessary to split the equation using the jacobian into a linear+ nonlinear part ?
 A: It is possible but it depends upon objectives. The point is that you would like solutions of the original ODE and the linearized ODE to 'look alike'. And for this they should really have the same linearized part (in the first sense).
'Look alike' may be interpreted as describing same qualitative behavior, notably stability of fixed points (but many other things may be of interest). Example in dimension one: 
$$ \frac{dy}{dt} = \sin(2y)-y . $$
$y=0$ is a fixed point and it is repulsive, i.e. if $y_0$ is any small initial value, then backward in time the solution $y(t)$ verifies $\lim_{t\rightarrow -\infty} y(t)=0$. This is seen by linearizing around the fixed point to
 $$ \frac{d \delta y}{dt} = 2\delta y - \delta y = \delta y $$
The solution is $\delta y(t) = y_0 e^t$ which indeed goes to zero as $t$ goes to $-\infty$. This implies the same for the genuine solution of the original equation. Indeed the set of solutions in a neighborhood of the fixed point are topologically conjugate (in this case also differentiably but this need not be the case in higher dimension).
If you made the suggested splitting and simply removed the $\sin$ then your 'linearized' equation would be: $\tilde{y}'= -\tilde{y}$ for which the solution is $\tilde{y}(t)=y_0 e^{-t}$ which has an attractive fixed point, precisely the opposite behavior of the real solution. 
An example where linearization  does not give the qualitative behaviour near the fixed point:
$ y' = y^3 $ where again $y=0$ is a fixed point and it is repulsive as before. But alas, this is not seen in the linearized equation when you throw away the $y^3$ term and the solution is a constant. So these solutions do not ressemble. Reason: The linearization matrix is the zero matrix and this case does not work. More generally, when you have a linearization matrix $M$ things work 'nicely' when there are no zero eigenvalue, better when there are no imaginary evals, and even better when evals are either in the right (repulsive fixed point) or the left complex half plane (attractive fixed point).
V.I. Arnold explains all this quite nicely in his book 'Ordinary differential equations' (and there are many other such books).
