For simplicity we write $G$ for $G^{\rm der}$ and $M_0$ for $M_0^{\rm der}$, then $G$ is a connected reductive group, and $M_0$ is a Levi subgroup of a minimal parabolic $P_0$ of $G$.
Let $$\xi\in{\rm ker}[H^1(k,M_0)\to H^1(k, G)]$$
be a cohomology class, represented by some 1-cocycle $c\in Z^1(k,M_0)$.
Since $\xi=[c]$ is contained in the kernel, the exists $g_1\in G(\bar k)$
such that $c_\sigma=g_1^{-1}\cdot \,^\sigma g_1$ for all 
$\sigma\in \Gamma:={\rm Gal}(\bar k/k)$.
We set 
$$M_1=g_1\cdot M_0\cdot g_1^{-1},\quad P_1=g_1\cdot P_0\cdot g_1^{-1},$$
then one can easily check that $M_1$ and $P_1$ are defined over $k$.
Since $(P_1,M_1)$ is conjugate to $(P_0,M_0)$ over $\bar k$, we see that $P_1$ is a parabolic of $G$ and that $M_1$ is a Levi subgroup of $P_1$. By Theorem 4.13(c) $P_0$ is conjugate to $P_1$ over $k$:
$$P_0=g\cdot P_1\cdot g^{-1}\quad\text{for some }g\in G(k).$$
Set $g_2=g g_1\in G(\bar k)$,
then
$$g_2 \cdot P_0 \cdot g_2^{-1}=P_0\,,$$
hence, $g_2\in N_G(P_0)=P_0$.
Now since $^\sigma g=g$, we notice that
$$c_\sigma=g_2^{-1}\cdot \,^\sigma g_2\,,$$
so we see that 
$$\xi\in{\rm ker}[H^1(k,M_0)\to H^1(k,P_0)].$$
Since $H^1(k,P_0)= H^1(k,M_0)$, see e.g. Lemma 1.13 in 
[Sansuc's paper](http://www.digizeitschriften.de/dms/img/?PID=GDZPPN002198746&physid=phys25#navi), we conclude that $\xi=1$, hence
$${\rm ker}[H^1(k,M_0)\to H^1(k, G)]=1.$$ A twisting argument shows that every fiber has only one element.