Let $\mathcal{B}(F)$ the algebra of all bounded linear operators on a complex Hilbert space $(F,\|\cdot\|)$. Let $M\in \mathcal{B}(F)^+$ (i.e. $\langle Mx\;, \;x\rangle\geq 0$ for all $x\in F$).
Let $S_1,S_2\in \mathcal{B}(F)$ such that $S_k(\text{ker}(M))\not\subset\text{ker}(M)$. Consider the follwoing subset of $\mathbb{C}^2$: \begin{eqnarray*} W &=&\{(\lambda_1,\lambda_2)\in \mathbb{C}^2;\;\exists(x_1,x_2)\in \text{ker}(M)\times \overline{\text{Im}(M)}\,;\|M^{1/2}x_2\|=1,\\ &&(\langle MS_1x_1,x_2\rangle+\langle MS_1x_2,x_2\rangle,\langle MS_2x_1,x_2\rangle+\langle MS_2x_2,x_2\rangle)=(\lambda_1,\lambda_2)\}. \end{eqnarray*}
I claim that $W=\mathbb{C}^2$. Is this claim true?
As far as I can tell, nothing you wrote prevents $S_1=S_2$, in which case $$W=\{(\lambda,\lambda):\ \lambda\in\mathbb C\}.$$
Example: take $F=\mathbb C^2$, $$ M=\begin{bmatrix}0&0\\0&1\end{bmatrix},\ \ S_1=S_2=\begin{bmatrix}0&0\\1&0\end{bmatrix}. $$ Then $$\ker M=\left\{\begin{bmatrix} \alpha\\0\end{bmatrix}:\ \alpha\in\mathbb C\right\},\ \ \ \text{Im}\,M=\left\{\begin{bmatrix}0\\ \beta\end{bmatrix}:\ \beta\in\mathbb C \right\}, $$ and $S_j\ker M=\text{Im}\,M$, $j=1,2$.
If $x_2=\begin{bmatrix} \gamma\\ \delta\end{bmatrix}$, then $\|M^{1/2}x_2\|=|\delta|$, so we need $\delta=1$. Now, with $x_1=\begin{bmatrix} r\\ s\end{bmatrix}$, $$ \langle MS_1x_1,x_2\rangle + \langle MS_1x_2,x_2\rangle =r+\gamma. $$ As $S_2=3S_1$, $$\langle MS_2x_1,x_2\rangle + \langle MS_2x_2,x_2\rangle=3\langle MS_1x_1,x_2\rangle + \langle MS_1x_2,x_2\rangle=3(r+\gamma).$$ Then $$ W=\{(r+\gamma,3(r+\gamma)):\ r,\gamma\in\mathbb C\}=\{(\lambda,3\lambda):\ \lambda\in\mathbb C\}\subsetneq\mathbb C^2. $$