I think what is going on is the following.  The confusion comes (as is often the case) from confusing "equals in a natural way" with "isomorphic".

As Meisam Soleimani Malekan says, for any Hilbert space $H$ with inner product $(\cdot|\cdot)$, and $\newcommand{\mc}{\mathcal}T\in\mc B(H)$, we have that $\ker T^*T=\ker T$, because $T^*T\xi=0\implies (T^*T\xi|\xi)=0 \implies \|T\xi\|^2=0\implies T\xi=0$.  Also, $\newcommand{\im}{\operatorname{Im}}(\im T)^\perp = \ker T^*$ by an easy calculation.

For your $F$ and $M\in\mc B(F)$ positive we hence have that $\ker M = \ker M^{1/2}$ and so on $F / \ker M$ we may define an inner product
$$ \newcommand{\ip}[2]{\langle{#1},{#2}\rangle} \ip{\overline{x}}{\overline{y}} = (Mx|y) = (M^{1/2}x|M^{1/2}y). $$
Here $\overline{x} = x+\ker M = x+\ker M^{1/2}$ is the equivalence class of $x\in F$.  It is evident that $\overline{x} = \overline{x'} \implies x-x'\in\ker M^{1/2} \implies M^{1/2}x = M^{1/2}x'$, so the definition is well-defined.

Alternatively, in any Hilbert space $H$, if $X\subseteq H$ is a closed subspace, then we can identify $H/X$ with $X^\perp$.  Indeed, let $P:H\rightarrow X^\perp$ be the orthogonal projection, and define $\theta:H/X\rightarrow X^\perp; \overline{\xi}\mapsto P(\xi)$.  This is well-defined, for $\overline{\xi}=\overline{\eta}\implies \xi-\eta\in X \implies P(\xi-\eta)=0$, and similarly $\theta$ is injective, as $P(\xi)=0$ exactly when $\xi\in X$.  By construction $\theta$ is onto, and so $\theta$ is an isomorphism.

**However** this construction, applied to your setting, identifies $F/\ker M$ with $(\ker M)^\perp = (\im M)^{\perp\perp} = \overline{\im M}$ which is already a Hilbert space, so complete.  The inner product we get on $F/\ker M$ is $(\overline x|\overline y) = (P(x)|P(y))$.  Suppose we choose $x$ and $y$ already in $\overline{\im M}$. Then $(\overline x|\overline y) = (P(x)|P(y)) = (x|y)$.  This **is not** the inner product you want.

**Instead** let's define $U : \overline{\im M} \rightarrow F/\ker M$ as follows.  Firstly notice that $\overline{\im M} = \overline{\im M^{1/2}}$ because $\overline{\im M} = (\ker M)^\perp = (\ker M^{1/2})^\perp = \overline{\im M^{1/2}}$.  Then define $U(\xi) = \overline{x}$ if $\xi = M^{1/2}x$ for some $x\in F$.  This is well-defined, for if $\xi=M^{1/2}y$ then $x-y\in\ker M^{1/2}=\ker M$ so $\overline{x}=\overline{y}$.  Furthermore, for $\xi=M^{1/2}x$ and $\eta=M^{1/2}y$,
$$ \ip{U(\xi)}{U(\eta)} = \ip{\overline x}{\overline y} = (Mx|y)
= (M^{1/2}x|M^{1/2}y) = (\xi|\eta). $$
Hence $U$ is an isometry.  We have only define $U$ on $\im M^{1/2}$ but as $U$ is an isometry it extends to $\overline{\im M^{1/2}}$.  Clearly $U$ is onto, and so $U$ is unitary, thus establishing the isomorphism between $\overline{\im M}$ and $F/\ker M$.