Strain compatibility From this link, strain compatibility for 2D problem with strains as
$$
   \varepsilon_{11} = \cfrac{\partial u_1}{\partial x_1} ~;~~
   \varepsilon_{12} = \cfrac{1}{2}\left[\cfrac{\partial u_{1}}{\partial x_2} + \cfrac{\partial u_{2}}{\partial x_1}\right]~;~~
   \varepsilon_{22} = \cfrac{\partial u_{2}}{\partial x_2} 
$$
is given by 
$$
   \cfrac{\partial^2 \varepsilon_{11}}{\partial x_2^2}
   - 2\cfrac{\partial^2 \varepsilon_{12}}{\partial x_1 \partial x_2}
   + \cfrac{\partial^2 \varepsilon_{22}}{\partial x_1^2} = 0
$$
Why the compatibility equation needs to have partial derivatives of strains?
Can we simply write the compatibility equation as 
$$
\varepsilon_{11} - 2\varepsilon_{12} + \varepsilon_{22}
$$
 A: The compatibility conditions are important to ensure the existence of a displacement field $u_i$ when a strain field $\varepsilon_{ij}$ is given. They are integrability conditions. You get them from the definitions of the strains:
$$
\varepsilon_{11}=u_{1,1}\\
\varepsilon_{12}=(u_{1,2}+u_{2,1})/2\\
\varepsilon_{22}=u_{2,2}
$$
With the first an last equation you can write $u_1=\int\varepsilon_{11}\ {\rm d}x_1$ and $u_2=\int\varepsilon_{22}\ {\rm d}x_2$. That you can insert in the definition of $\varepsilon_{12}$,
$$
\varepsilon_{12}=\left((\int\varepsilon_{11}\ {\rm d}x_1)_{,2}+(\int\varepsilon_{22}\ {\rm d}x_2)_{,1}\right)/2.
$$
Of this expression you take a derivative w.r.t. $x_1$ and $x_2$ to get
$$
\varepsilon_{12,12}=\left(\varepsilon_{11,22}+\varepsilon_{22,11}\right)/2.
$$
which is the equation in your question. If this does not hold then you get different $u_1$ when 


*

*integrating $\varepsilon_{11}$ or 

*when integrating first $\varepsilon_{22}$, inserting $u_2$ from that into $\varepsilon_{12}$ and then integrating for $u_1$. 


Now you asked why we would not just take a linear combination of the strains as compatibility condition. The answer is: It does not say anything about the integrability. 
