Let $C(x)$ a linear function on $\mathbb{R}$.
Then we have:
$$ \begin{align*} C\left(S_0^1\right)&=C\left( \frac{y_m-r}{y_m-y_1}S_0^1+\frac{r-y_1}{y_m-y_1}S_0^1 \right) \\ &=\frac{y_m-r}{y_m-y_1}C\left( \frac{1+y_1}{1+r}S_0^1 \right)+\frac{r-y_1}{y_m-y_1}C \left( \frac{1+y_m}{1+r}S_0^1 \right), \end{align*} $$
for $S_0^1>0,y_1<0<r<y_m$.
The first equality follows by linearity of $C$ and
$$\frac{y_m-r}{y_m-1},\frac{r-y_1}{y_m-y_1}>0 \, \operatorname{with} \,\frac{y_m-r}{y_m-y_1}+\frac{r-y_1}{y_m-y_1}=1.$$
(some weighted average).
Question: Why holds the second equality?
Intuitively I think we have a weighted average of some convex combination of a linear function, but I am not able to formally understand it, maybe you can show it.
For $y_m=-y_1$ the second equality is true as it adds to zero, but the general case $y_1<0<r<y_m$ I could not show.
It would be here sufficient to show, since $C$ is linear:
$ \frac{y_m-r}{y_m-y_1}\frac{1+y_1}{1+r}+\frac{r-y_1}{y_m-y_1}\frac{1+y_m}{1+r}=1 $
It may be that the condition $y_1=-y_m$ was forgotten, in which case you may easily contradict the equality by showing some fit numerical counterexample for $y_1\neq-y_m$ with $y_1<0<r<y_m$...
Note that $$(y_m-r)(1+y_1)+(r-y_1)(1+y_m)=y_m+y_1y_m-r-ry_1+r+ry_m-y_1-y_1y_m\\=y_m-ry_1+ry_m-y_1=y_m(1+r)-y_1(1+r)=(y_m-y_1)(1+r)$$ so $${y_m-r\over y_m-y_1}{1+y_1\over1+r}+{r-y_1\over y_m-y_1}{1+y_m\over1+r}=1$$ Q.E.D.