I think what is going on is the following.

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.  Notice that we have not established that $F/\ker M$ is complete for this inner-product.

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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.

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**Instead** let's define $U:\im(M^{1/2})\rightarrow F/\ker M$ as follows.  Define $U(\xi) = \overline{x}$ if $\xi = M^{1/2}x$ for some $x\in F$.  This is well-defined, for if also $\xi=M^{1/2}y$ 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.  Clearly $U$ is surjective.  So
$$ U^{-1} : \big( F/\ker M, \ip{\cdot}{\cdot} \big) \rightarrow
\big( \im(M^{1/2}), (\cdot|\cdot) \big) $$
is an isometric linear isomorphism.  This identifies $F/\ker M$, given the inner-product you have defined, with the (in general not closed) subspace $\im(M^{1/2})$ of $F$.  In particular, $F/\ker M$ might fail to be a Hilbert space.

However, if we take the _completion_ of $F/\ker M$ then $U^{-1}$ extends to a unitary showing that $\overline{F/\ker M}$ is isomorphic to $\overline{\im(M^{1/2})} = \overline{\im(M)}$, the latter viewed as a subspace of $F$.

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Your theorem  suggests that we give $\im(M^{1/2})$ a _different_ inner-product,
$$ (M^{1/2}x|M^{1/2}y)_0 = (Px|Py) $$
where I have written $(\cdot|\cdot)_0$ to avoid confusion with $(\cdot|\cdot)$ which is the given inner-product on $F$.  Define a _different_ map $V:F/\ker M \rightarrow \im M^{1/2}$ by
$$ V(\overline x) = Mx = M^{1/2} M^{1/2} x \in \im M^{1/2}. $$
Again, this is well-defined.  As $\overline{\im M} = (\ker M)^\perp = (\ker M^{1/2})^\perp = \overline{\im M^{1/2}}$, we have that $PM^{1/2}x=M^{1/2}x$, and so
$$ (Mx|My)_0 = (M^{1/2} M^{1/2} x|M^{1/2} M^{1/2} y)_0 = (PM^{1/2}x|PM^{1/2}y)
= (M^{1/2}x|M^{1/2}y) = \ip{\overline x}{\overline y}. $$
Thus
$$ V:\big( F/\ker M, \ip{\cdot}{\cdot} \big) \rightarrow \big( \im M^{1/2}, (\cdot|\cdot)_0 \big) $$
is an isometry, as you want.