Let $\mathcal{B}(F)$ the algebra of all bounded linear operators on a complex Hilbert space $(F,\|\cdot\|)$. Let $M\in \mathcal{B}(F)$ be a positive semidefinite operator.
The normed space $(\operatorname{Im}(M^{1/2}),\|\cdot\|_{\operatorname{Im}(M^{1/2})})$ is Hilbert, where $$\|M^{1/2}x\|_{\operatorname{Im}(M^{1/2})}=\|P_{\overline{\operatorname{Im}(M)}}x\|,\;\forall\, x\in F$$
If $S$ is a bounded operator on $\operatorname{Im}(M^{1/2})$, then we can prove that $$\|S\|=\sup_{\substack{x\in \overline{\operatorname{Im}(M^{1/2})},\\ x\not=0}} \frac{\|SM^{1/2}x\|_{\operatorname{Im}(M^{1/2})}}{\|M^{1/2}x\|_{\operatorname{Im}(M^{1/2})}}.$$
I want to prove that $$\sup_{\substack{x\in \overline{\operatorname{Im}(M^{1/2})},\\ x\not=0}} \frac{\|SM^{1/2}x\|_{\operatorname{Im}(M^{1/2})}}{\|M^{1/2}x\|_{\operatorname{Im}(M^{1/2})}}=\sup_{\substack{x\in \overline{\operatorname{Im}(M)},\\ x\not=0}} \frac{\|SMx\|_{\operatorname{Im}(M^{1/2})}}{\|Mx\|_{\operatorname{Im}(M^{1/2})}}.$$
Let's prove $\overline{\operatorname{Im} M} = \overline{\operatorname{Im} M^{1/2}}$.
For $y = Mx \in \operatorname{Im}M$ we have $y = Mx = M^{1/2}(M^{1/2}x) \in \operatorname{Im}M^{1/2}$.
Hence $\operatorname{Im}M \subseteq \operatorname{Im}M^{1/2}$. By taking the closures it follows $\overline{\operatorname{Im}M} \subseteq \overline{\operatorname{Im}M^{1/2}}$.
For the converse inclusion, you can see that every $y = M^{1/2}x \in \operatorname{Im} M^{1/2}$ can be expressed as a limit
$$y = Mx = \lim_{n\to\infty} Mx_n$$
of a sequence $(Mx_n)_n$ in $\operatorname{Im} M$ so we conclude $\operatorname{Im} M^{1/2} \subseteq \overline{\operatorname{Im} M}$. By taking the closures we conclude $\overline{\operatorname{Im} M^{1/2}} \subseteq \overline{\operatorname{Im} M}$.
Let $x \in \overline{\operatorname{Im} M}$, $x \ne 0$ be arbitrary. Define $y = M^{1/2}x \in \operatorname{Im} M^{1/2}$. Since $x \in \overline{\operatorname{Im} M} = \overline{\operatorname{Im} M^{1/2}}$ certainly $x \perp \ker M^{1/2}$ and $x \ne 0$ so $y = M^{1/2}x \ne 0$. We have
$$\frac{\|SMx\|_{\operatorname{Im} M^{1/2}}}{\|Mx\|_{\operatorname{Im} M^{1/2}}} = \frac{\|SM^{1/2}y\|_{\operatorname{Im} M^{1/2}}}{\|M^{1/2}y\|_{\operatorname{Im} M^{1/2}}}$$
Therefore
$$\left\{\frac{\|SMx\|_{\operatorname{Im} M^{1/2}}}{\|Mx\|_{\operatorname{Im} M^{1/2}}} : x \in \overline{\operatorname{Im} M}, x \ne 0\right\}\subseteq\left\{\frac{\|SM^{1/2}x\|_{\operatorname{Im} M^{1/2}}}{\|M^{1/2}x\|_{\operatorname{Im} M^{1/2}}} : x \in \overline{\operatorname{Im} M^{1/2}}, x \ne 0\right\}$$
For the converse inclusion, let $y \in \overline{\operatorname{Im} M^{1/2}}$, $y \ne 0$. There exists a sequence $(y_n)_n$ in $\operatorname{Im} M^{1/2}$ such that $y_n \xrightarrow{n\to\infty} y$. We can assume that $y_n \ne 0, \forall n \in \mathbb{N}$. There exists a sequence $(x_n)_n$ in $F$ such that $y_n = M^{1/2}x, \forall n \in \mathbb{N}$. We can choose $x_n$ to be in $\overline{\operatorname{Im} M^{1/2}}$ because of the decomposition $F = \overline{\operatorname{Im} M^{1/2}} \oplus \ker M^{1/2}$. We then also have $x_n \in \overline{\operatorname{Im} M}$ and $x_n \ne 0$ for all $n \in \mathbb{N}$.
Now we have
\begin{align} \frac{\|SM^{1/2}y\|_{\operatorname{Im} M^{1/2}}}{\|M^{1/2}y\|_{\operatorname{Im} M^{1/2}}} &= \frac{\|SM^{1/2}(\lim_{n\to\infty}y_n)\|_{\operatorname{Im} M^{1/2}}}{\|M^{1/2}(\lim_{n\to\infty}y_n)\|_{\operatorname{Im} M^{1/2}}}\\ &= \frac{\lim_{n\to\infty}\|SM^{1/2}y_n\|_{\operatorname{Im} M^{1/2}}}{\lim_{n\to\infty}\|M^{1/2}y_n\|_{\operatorname{Im} M^{1/2}}}\\ &= \lim_{n\to\infty} \frac{\|SM^{1/2}y_n\|_{\operatorname{Im} M^{1/2}}}{\|M^{1/2}y_n\|_{\operatorname{Im} M^{1/2}}}\\ &= \lim_{n\to\infty} \frac{\|SM^{1/2}M^{1/2}x_n\|_{\operatorname{Im} M^{1/2}}}{\|M^{1/2}M^{1/2}x_n\|_{\operatorname{Im} M^{1/2}}}\\ &= \lim_{n\to\infty} \frac{\|SMx_n\|_{\operatorname{Im} M^{1/2}}}{\|Mx_n\|_{\operatorname{Im} M^{1/2}}} \in \overline{\left\{\frac{\|SMx\|_{\operatorname{Im} M^{1/2}}}{\|Mx\|_{\operatorname{Im} M^{1/2}}} : x \in \overline{\operatorname{Im} M}, x \ne 0\right\}} \end{align}
because each of $\frac{\|SMx_n\|_{\operatorname{Im} M^{1/2}}}{\|Mx_n\|_{\operatorname{Im} M^{1/2}}}$ is an element of $\left\{\frac{\|SMx\|_{\operatorname{Im} M^{1/2}}}{\|Mx\|_{\operatorname{Im} M^{1/2}}} : x \in \overline{\operatorname{Im} M}, x \ne 0\right\}$.
We conclude
$$\left\{\frac{\|SM^{1/2}x\|_{\operatorname{Im} M^{1/2}}}{\|M^{1/2}x\|_{\operatorname{Im} M^{1/2}}} : x \in \overline{\operatorname{Im} M^{1/2}}, x \ne 0\right\} \subseteq \overline{\left\{\frac{\|SMx\|_{\operatorname{Im} M^{1/2}}}{\|Mx\|_{\operatorname{Im} M^{1/2}}} : x \in \overline{\operatorname{Im} M}, x \ne 0\right\}}$$
So we have the equality of closures
$$\overline{\left\{\frac{\|SM^{1/2}x\|_{\operatorname{Im} M^{1/2}}}{\|M^{1/2}x\|_{\operatorname{Im} M^{1/2}}} : x \in \overline{\operatorname{Im} M^{1/2}}, x \ne 0\right\}} = \overline{\left\{\frac{\|SMx\|_{\operatorname{Im} M^{1/2}}}{\|Mx\|_{\operatorname{Im} M^{1/2}}} : x \in \overline{\operatorname{Im} M}, x \ne 0\right\}}$$
Finally
\begin{align} \|S\|_{\mathcal{B}(\operatorname{Im}M^{1/2})} &= \sup \left\{\frac{\|SM^{1/2}x\|_{\operatorname{Im} M^{1/2}}}{\|M^{1/2}x\|_{\operatorname{Im} M^{1/2}}} : x \in \overline{\operatorname{Im} M^{1/2}}, x \ne 0\right\} \\ &= \sup\overline{\left\{\frac{\|SM^{1/2}x\|_{\operatorname{Im} M^{1/2}}}{\|M^{1/2}x\|_{\operatorname{Im} M^{1/2}}} : x \in \overline{\operatorname{Im} M^{1/2}}, x \ne 0\right\}}\\ &= \sup \overline{\left\{\frac{\|SMx\|_{\operatorname{Im} M^{1/2}}}{\|Mx\|_{\operatorname{Im} M^{1/2}}} : x \in \overline{\operatorname{Im} M}, x \ne 0\right\}}\\ &= \sup \left\{\frac{\|SMx\|_{\operatorname{Im} M^{1/2}}}{\|Mx\|_{\operatorname{Im} M^{1/2}}} : x \in \overline{\operatorname{Im} M}, x \ne 0\right\} \end{align}
because for any nonempty $A \subseteq \mathbb{R}$ we have $\sup A= \sup \overline{A}$.