Suppose that $r'(\theta)>0$. The following figure should make clear that saying $r(\theta)\leq w$ is equivalent to saying that $\theta\leq r^{-1}(w)$ (where $r^{-1}(\cdot)$ is the inverse of $r$):
So we can rephrase your question as 'why should $E(\theta|\theta<r^{-1}(w))$ be non-linear?'
Let's calculate this expectation (assuming that $\theta$ is distributed on support $[\underline{\theta},\overline{\theta}]$ according to CDF $F$):
$$E(\theta|\theta<r^{-1}(w))=\frac{1}{F[r^{-1}(w)]}\int_\underline{\theta}^{r^{-1}(w)}\!\theta F'(\theta)d\theta.$$
For $E(\theta|\theta<r^{-1}(w))$ to be linear in $w$ we would need $\partial E(\theta|\theta<r^{-1}(w))/\partial w$ to be constant.
Differentiating with respect to $w$ we have
$$\frac{\partial E(\theta|\theta<r^{-1}(w))}{\partial w}=\frac{F'[r^{-1}(w)]\frac{\partial r^{-1}(w)}{\partial w}\left[F[r^{-1}(w)]r^{-1}(w)-\int_\underline{\theta}^{r^{-1}(w)}\!\theta F'(\theta)d\theta\right]}{F[r^{-1}(w)]^2}$$
The above expression isn't the prettiest thing to look at, but a quick glance should convince you that it will only be constant under very special circumstances. For example, even if we take a highly linear environment like $r(\theta)=\theta$, $F(\theta)=\theta$, we have
$$\frac{\partial E(\theta|\theta<r^{-1}(w))}{\partial w}=\frac{w^2+\underline{\theta}^2}{2w^2},$$
so $E(\theta|\theta<r^{-1}(w))$ is non-linear.
