Let us restate the problem for ease of reference.
The equations are

$\dot{x}(t)=y(t)$    
$\dot{y}(t)= - 4 x(t) + y(t)^2$

Physically they describe an anharmonic oscillator with spatial coordinate $x(t)$ and velocity $y(t)$ in the potential

$U = 2 x^2 - \frac{1}{3} x^3$

and the energy

$E = \frac{1}{2} y^2 + U(x)$

The problem is then: given two solutions 

$s1(t) = (x1(t), y1(t))$     
$s2(t) = (x2(t), y2(t))$ 

corresponding to the initial conditions 

$s1(0) = (x1(t=0) = x10, y1(t=0) = y10)$
$s2(0) = (x2(t=0) = x20, y2(t=0) = y20)$

Assuming that $y10$ and $y20$ do not vanish show that the quantity

$d(t) = y1(t) + y2(t)$

becomes $\neq 0$ for times $t>0$ even if it is $= 0$ for $t = 0$, i.e.

$y10 + y20 = 0$.

We prove it indirectly, assuming $d(t) = 0$ for all times.

The idea is to expand the solution $y(t)$ into a power series in t.

$y(t) = y(0) + t \dot{y}(0) + \frac{1}{2} \ddot{y}(0) + \frac{1}{3} \frac{\partial ^3y(0)}{\partial t^3} + ...$

In order to have $d(t) = 0$, all coefficients must vanish. These can be expressed through the initial values, and the first coefficients are

$c(0) = (y10 + y20)  $       
$c(1) =  - 4 (x10 + x20) + (x10^2 + x20^2)  $       
$c(2) =  - 4 (y10 + y20) + 2 ( x10 y10 + x20 y20 ) $    
$c(3) =  2 (y10^2 + y20^2) + 2 ( x10^3+x20^3 ) -12 (x10^2 + x20^2) + 16 (x10 + x20)) $ 

Since we have $y10 + y20 = 0$, $d(t) = 0$ requires
      
$c(1) = 0 = - 4 (x10 + x20) + (x10^2 + x20^2)  $       
$c(2) = 0 = 2 y10 (x10 - x20) $    
$c(3) = 0 = 4 y10^2 + 2 ( x10^3+x20^3 ) -12 (x10^2 + x20^2) + 16 (x10 + x20)) $

From $c(2) = 0$ and $y10 \neq 0$ we find $x20 = x10$. Hence

$c(1) = 0 = - 4 x10 + x10^2  $         
$c(3) = 0 =  y10^2 +  x10(x10-2)(x10-4) $

Both solutions of $c(1) = 0$ give $y10 = 0$. The contradiction proves the assertion of the OP.

Observation: if we would drop the term x^2 in the second equation we can have $d(t) = 0$ by taking $x10 = x20$