The ordinary wave equation
$$\frac{\partial^2 u}{\partial t^2} = v_w^2 \frac{\partial^2 u}{\partial x^2}$$
where $u$ is the wave function and $v_w$ is the wave speed, is linear (in the sense of a linear operator - or "homogeneous and linear" in the sense of differential equations), in that if $u_1$ and $u_2$ - both functions of time as well as space - are solutions, then any weighted combination, i.e. $u = a_1 u_1 + a_2 u_2$, for constants $a_1$ and $a_2$, and that is to say, whose values are given by $u(t, x) = a_1 u_1(t, x) + a_2 u_2(t, x)$, is also a solution. This can be seen by noting that each side consists only of differentiations and no products, so that the expressions will "distribute" nicely over the sum.
In general relativity, if you have two metric waves $g_1$ and $g_2$, now tensor-valued instead of scalar-valued as above (so a "tensor wave", c.f. how EM waves are "vector waves"), while $g = a_1 g_1 + a_2 g_2$ may be another space-time metric tensor field, it is not a solution of the corresponding nonlinear wave equation, meaning that while "at any given spatial slice" it may be a valid instantaneous configuration of the geometry field, the propagation is all wrong.
That is because that the wave equations for gravitational waves are nonlinear wave equations. A simple example of such a wave equation would be
$$\frac{\partial^2 u}{\partial t^2} + v_w^2 \frac{\partial^2 u}{\partial x^2} + k u^2$$
for some $k > 0$. If you have $u_1$ and $u_2$ as solutions, then $u = u_1 + u_2$ is not a solution, for while at any given time it may be okay as a configuration of the wave system, the future evolution of that configuration is no longer merely the sum of the evolutions of the components - the squaring part introduces cross terms between the two, namely $2 u_1 u_2$, which cannot be created simply by taking the time derivative on the left, for it just distributes right across the sum, and introduces no new cross terms.
More specifically, the Einstein field equations are an equation in the Ricci tensor $R_{\mu \nu}$, and each component there takes (lots and lots of) products of partial derivatives of the metric tensor components. Hence they are all cross terms.
As to why this nonlinearity exists, you should note that even from elementary calculus, curvature of a simple curve given by a differentiable real valued function of one real variable is a nonlinear property in that simply adding the two curves won't add the curvatures. The Ricci tensor is a curvature for a general manifold of any dimensionality and not just such a curve, so it should be expected likewise to be similarly nonlinear in the "shape" thereof (here given by the metric field).