Prove that we can't find effective bounds on the point guaranteed by the Mean Value Theorem.

First, imagine $f$ is some such polynomial for $\xi$. Then let $g(x)=f(x)-Mx$. We have $g(0)=g(1)=0$, and $\xi$ is the unique number in $(0,1)$ where $g'(x)=0$. So allow me to replace the problem as written with the Mean Value Theorem to one about Rolle's Theorem.

Below is a proof that if you have $\xi>\frac{1}{2}$, take some integer $n>\frac{1-2\xi}{\xi-1}$, and then take $t=\frac{(n+2)\xi^2-(n+1)\xi}{-2\xi+1}$. Then the polynomial $g(x)=(x+t)^nx(1-x)$ satisfies $g(0)=g(1)=0$, and there is a unique number in $(0,1)$ where $g'(x)=0$, and that number is $\xi$.

If $\xi<\frac{1}{2}$, there is a symmetric construction with $t<-1$. And if $\xi=\frac{1}{2}$, just take $g(x)=x(1-x)$.

For example, with $\xi=\frac{e}{\pi}$, we can take $n=6$, and $t=\frac{8(e/\pi)^2-7(e/\pi)}{-2(e/\pi)+1}\approx0.09233\ldots$. Then $g(x)=(x+t)^6x(1-x)$ is such that $g'$ has only one zero in $(0,1)$, and it is located at $\frac{e}{\pi}$. See this demonstrated at WolframAlpha.

Explanation

Assume $\xi>\frac{1}{2}$. Consider $g(x)=(x+t)^nx(1-x)$ for $n\in\mathbb{N}$ and $t\in\mathbb{R}_{\gt0}$. Then $$ \begin{align} g'(x)&=n(x+t)^{n-1}x(1-x)+(x+t)^n(1-x)-(x+t)^nx\\ &=(x+t)^{n-1}\big(nx(1-x)+(x+t)(1-x)-(x+t)x\big)\\ &=(x+t)^{n-1}\big(x^2(-n-2)+x(n+1-2t)+t\big)\\ \end{align} $$ The zeros of $g'$ are $-t$ (which is not in $(0,1)$) and $$\frac{-(n+1-2t)\pm\sqrt{(n+1-2t)^2+4(n+2)t}}{-2(n+2)}=\frac{A\pm B}{C}$$ Since $n,t>0$, it follows that $|B|>|A|$. It follows that one of these two roots is negative (so not in $(0,1)$) and the other is positive. So if the positive root is equal to $\xi$, then $g$ satisfies the Rolle's version of the proposition. We have freedom to choose $n\in\mathbb{N}$, $t\in\mathbb{R}_{\gt0}$, so maybe we can choose them well. In the following, we attempt to solve for $t$ in terms of $\xi$ and $n$.

$$\begin{align} \xi&=\frac{-(n+1-2t)\pm\sqrt{(n+1-2t)^2+4(n+2)t}}{-2(n+2)}\\ -2(n+2)\xi&=-(n+1-2t)\pm\sqrt{(n+1-2t)^2+4(n+2)t}\\ -2(n+2)\xi+n+1-2t&=\pm\sqrt{(n+1-2t)^2+4(n+2)t} \end{align}$$ Squaring both sides: $$\begin{align} [-2(n+2)\xi+n+1]^2-4t[-2(n+2)\xi+n+1]+4t^2&=(n+1-2t)^2+4(n+2)t\\ [-2(n+2)\xi+n+1]^2-4t[-2(n+2)\xi+n+1]+4t^2&=(n+1)^2-4(n+1)t+4t^2+4(n+2)t\\ [-2(n+2)\xi+n+1]^2-4t[-2(n+2)\xi+n+1]&=(n+1)^2+4t\\ [-2(n+2)\xi+n+1]^2-(n+1)^2&=4t[-2(n+2)\xi+n+2]\\ 4(n+2)^2\xi^2-4(n+2)(n+1)\xi&=4t(-2(n+2)\xi+n+2)\\ \end{align}$$ $$\begin{align} t&=\frac{4(n+2)^2\xi^2-4(n+2)(n+1)\xi}{4(-2(n+2)\xi+n+2)}\\ &=\frac{(n+2)\xi^2-(n+1)\xi}{-2\xi+1} \end{align}$$ We have assumed $\xi>\frac{1}{2}$, so the denominator is negative. We need $t$ to be positive, so we need the numerator to be negative. Can we choose $n$ to make that happen? $$\begin{align} (n+2)\xi^2-(n+1)\xi&<0\\ n(\xi^2-\xi)&<\xi-2\xi^2\\ n(\xi-1)&<1-2\xi\\ n&>\frac{1-2\xi}{\xi-1} \end{align}$$ So yes. If you have $\xi>\frac{1}{2}$, take some integer $n>\frac{1-2\xi}{\xi-1}$, and then take $t=\frac{(n+2)\xi^2-(n+1)\xi}{-2\xi+1}$. Then the polynomial $g(x)=(x+t)^nx(1-x)$ satisfies $g(0)=g(1)=0$, and there is a unique number in $(0,1)$ where $g'(x)=0$, and that number is $\xi$.

And restating from the introduction, if $\xi<\frac{1}{2}$, there is a symmetric construction where $t<-1$. And if $\xi=\frac{1}{2}$, just take $g(x)=x(1-x)$.

This approach was motivated by starting with $x(1-x)$, and then multiplying by some power of $(x+t)$ that would "warp" the parabola in between $0$ and $1$ without making it wiggle. Something that would stretch $x(1-x)$ vertically, but moreso on the right side than the left side, to push the extremum farther to the right. (Or the other way when $\xi<\frac{1}{2}$.)

Just a few words on solving the nonpolynomial problem: First take $M=0.$ Then $f(x)=x(1-x)$ is a solution for $\xi=1/2.$ For other values of $\xi\in (0,1),$ we look at $g_p(x)=f(x^p)$ for $p>0.$ We have $g_p'(x)=px^{p-1}f'(x^p)=0$ iff $x=(1/2)^{1/p}.$ Thus if $\xi$ is given, we take $p=\ln(1/2)/\ln \xi,$ and $g_p$ solves the problem. Finally, if $M\ne 0$ and $\xi$ is given, we take the same $p$ and the function $M(x+g_p(x))$ solves the problem.

On to polynomial solutions: WLOG $M=0$, for we can use the same idea as above for $M\ne 0.$

Let $f(x)=x(1-x)$ as above. If $p$ is a polynomial with $p(0)=0,$ $p(1)=1$ and $p'>0$ on $(0,1),$ then $f\circ p$ is a polynomial that solves the problem for for the unique value $\xi \in(0,1)$ such that $p(\xi)=1/2.$

For $0\le b \le 1,$ set $p_b(x)=(1-b)x^2 +bx.$ Then $p_b(0)=0,p_b(1)=1,$ and $p_b'>0$ on $(0,1).$ Consider the equation $p_b(x)-1/2=0.$ If $b=1,$ then $x=1/2$ is a solution. For $b\in [0,1)$ we have a quadratic equation whose solution in $[0,1]$ is

$$x= \frac{(b^2+2(1-b))^{1/2}-b}{2(1-b)}.$$

Verify the right side, as a function of $b,$ strictly decreases on $[0,1)$ from $1/2^{1/2}$ to $1/2.$ Thus for $\xi\in [1/2,1/2^{1/2}],$ there is a unique $b_{\xi}\in [0,1]$ such that $p_{b_{\xi}}(\xi)=1/2.$ Verify that $b_{\xi}$ is given by the formula

$$b_{\xi} = \frac{1/2-\xi^2}{\xi(1-\xi)}.$$

So we've solved the problem for $\xi\in[1/2,1/2^{1/2}].$ But we've also solved it for $\xi\in [1/2^{1/2},1/2^{1/4}].$ Just check that that for such $\xi,$ $f\circ p_{b_{\xi^2}}(x^2)$ does the job. We can keep moving to the right with such intervals and their solutions. We thus obtain solutions for all $\xi\in [1/2,1).$

What about $\xi\in (0,1/2]?$ That's easy, now that we've handled the other side. Just check that if $g$ is a solution for $\xi\in [1/2,1),$ then $1-g(1-x)$ is a solution for $1-\xi.$