First, for a Dirac operator $D$, it extends to a Fredholm operator
\begin{equation}
D^+\colon H^1(M,E^+)\to L^2(M,E^-),
\end{equation}
see Lawson-Michelsohn "Spin Geometry", Page 193, Theorem 5.2. 

Also, a classical result in functional analysis says the index of the Fredholm operator is locally constant, see for instance Lemma 16.18 of https://ocw.mit.edu/courses/18-965-geometry-of-manifolds-fall-2004/8a7e4dd837d1bdd6988e0330babb8c5e_lecture16_17.pdf

Therefore, using the above lemma for the family $t\in [0,1]\mapsto tD_0^++(1-t)D_1^+$, the index is constant as required.

To see that the Fredholm index
\begin{equation}
Ind(D^+)=\dim \mathrm{kernel}(D^+)-\mathrm{codim\ range}(D^+)
\end{equation}
coincides with the one you want, we only need to prove that
\begin{equation}
\mathrm{codim\ range}(D^+)=\dim \ker(D^-).
\end{equation}
Now we consider the dual $D^- $ of $D^+$
\begin{equation}
D^-\colon L^2(M,E^-)\to H^{-1}(M,E^+),
\end{equation}
Using Rudin Functional Analysis Theorem 4.12, we get
\begin{equation}
L^2(M,E^-)=\ker D^-\oplus \overline{\mathrm{range}(D^+)},
\end{equation}
and since $D^+$ is Fredholm, its image is closed, $\overline{\mathrm{range}(D^+)} =\mathrm{range}(D^+)$, so
\begin{equation}
L^2(M,E^-)=\ker D^-\oplus\mathrm{range}(D^+),
\end{equation}
from which we clearly get $\mathrm{codim\ range}(D^+)=\dim \ker(D^-)$ as required. 

A subtle point is to distinguish different adjoints, first,
\begin{equation}
(D^+)^*\colon L^2(M,E^-)\to H^{1}(M,E^+),
\end{equation}
is defined by
\begin{equation}
((D^+)^*s,t)_{H^{1}(M,E^+)}=(s,D^+t)_{L^2(M,E^-)}
\end{equation}
for $s\in L^2(M,E^-),t\in H^{1}(M,E^+)$. But remember that this adjoint is NOT the way we define $D^-$, because $D^-$ is defined via the $L^{2}(M,E^+)$ product, not the $H^{1}(M,E^+)$ product:
\begin{equation}
((D^+)^*s,t)_{L^{2}(M,E^+)}=(s,D^+t)_{L^2(M,E^-)},
\end{equation}
that is why $(D^+)^*s$ is in $H^{-1}(M,E^+)$, not $H^{1}(M,E^+)$.