First, for a matrix $M$ the usual notation for for the image is $\operatorname{Im}(M)$.
Second remark is that $\dim\ker(M)+\dim\operatorname{Im}(M)=n$, not $\ker(M)+\operatorname{Im}(M)=n$.
As to your question, since $M$ is self-adjoint, $M=M^*$ it indeed holds that $\operatorname{Im}(M)=\operatorname{Im}(M^*)$, but the only thing it proves is that $\dim\ker(M^*)+\dim\operatorname{Im}(M)=n$, but this, by itself does not prove that $\ker(M^*)+\operatorname{Im}(M)=V$ (where $V=F^n$). To prove that $\ker(M^*)$ is a complement of $\operatorname{Im}(M)$ you need to prove further that $\ker(M^*)\cap\operatorname{Im}(M)=\{0\}$. Then it will follow that: $\begin{align*}\dim\left(\ker(M^*)+\operatorname{Im}(M)\right)&=\dim\ker(M^*)+\dim\operatorname{Im}(M)-\dim\left(\ker(M^*)\cap\operatorname{Im}(M)\right)\\ &=n-0=0\end{align*}$ And hence $\ker(M^*)+\operatorname{Im}(M)=V$.
As for your second question, prove by double-inclusion. Clearly, $\ker(M)\subseteq\ker(M^*M)$ (do you see why?). Now take any $v\in\ker(M^*M)$. Then $M^*Mv=0$, so $Mv\in\ker(M^*)$. Then $Mv\in\ker(M^*)\cap\operatorname{Im}(M)$. It follows that $Mv=0$ (why?). Hence $v\in\ker(M)$. Do the same for the second part as well.