It's not true in general that $\widetilde{\mathcal F(X)}=\mathcal F$; there exist $\mathcal O_X$-modules which are not of the form $\widetilde M$ for any $M$.
However, what we do know is that there is a restriction map $\mathcal F(X)\to\mathcal F(X_f)$ for any $f\in A$. But $\mathcal F(X_f)$ is a module over $\mathcal O_X(X_f)=A_f$, so multiplication by $f$ is an automorphism of $\mathcal F(X_f)$, so by the universal property of localization of modules there is an induced ($A_f$-module) morphism $\mathcal F(X)_f\to\mathcal F(X_f)$ (making the corresponding triangle commute).
Combining with the map $M_f\to \mathcal F(X)_f$ you found, we get a map $\widetilde M(X_f)=M_f\to\mathcal F(X_f)$, and these you can glue to get a sheaf morphism $\widetilde M\to\mathcal F$.
$\newcommand{\Spec}{\mathrm{Spec}}$$\newcommand{\Hom}{\mathrm{Hom}}$$\newcommand{\Q}{\mathbb{Q}}$$\newcommand{\Z}{\mathbb{Z}}$
You are trying to specify a sheaf homomorphism $\underline{\mathbb{Z}/3\mathbb{Z}}\to \mu_3$ over $U_1:=\mathrm{Spec}(\Q[x]/(x^2+x+1))$ for the choice of the element $x\in \mu_3(U_1)$.
By definition this means that for every etale map $U\to U_1$ we have a map of abelian groups $\underline{\mathbb{Z}/3\mathbb{Z}}(U)\to \mu_3(U)$ such that for any etale map $V\to U$ we have that the diagram
$$\begin{matrix}\underline{\Z/3\Z}(U) & \to & \mu_3(U)\\ \downarrow & & \downarrow\\ \underline{\Z/3\Z}(V) & \to & \mu_3(V)\end{matrix}$$
commutes.
You were then confused because it seemed like if we set $U_2:=\Spec(\Q[x,y]/(x^2+x+1,y^2+y+1))$ that there is ambiguity in the map
$$(\Z/3\Z)^2=\underline{\Z/3\Z}(U)\to \mu_3(U)$$
But, note that
$$U_2=V_1\sqcup V_2\cong U_1\sqcup U_1$$
essentially because
$$\Q[x,y]/(x^2+x+1,y^2+y+1)\cong (\Q[x]/(x^2+x+1))[y]/(y-x)\times (\Q[x]/(x^2+x+1))[y]/(y-x^2)$$
and where we set
$$V_1:=\Spec((\Q[x]/(x^2+x+1))[y]/(y-x)),\qquad V_2:=\Spec((\Q[x]/(x^2+x+1))[y]/(y-x^2))$$
So, from our compatability conditions we see that the map $\underline{\Z/3\Z}(U_2)\to \mu_3(U_2)$ is actually determined by the two maps
$$\underline{\Z/3\Z}(V_1)\to \mu_3(V_1),\qquad \underline{\Z/3\Z}(V_2)\to \mu_3(V_2)$$
But, since we have the commutativity of the diagrams
$$\begin{matrix}\underline{\Z/3\Z}(U_1) & \to & \mu_3(U_1)\\ \downarrow & & \downarrow\\ \underline{\Z/3\Z}(V_i) & \to & \mu_3(V_i)\end{matrix}$$
and the vertical maps are isomorphisms, we see that the map
$$\underline{\Z/3\Z}(V_1)\to \mu_3(V_1)$$
sends $1$ to $x=y$ and the map
$$\underline{\Z/3\Z}(V_2)\to \mu_3(V_2)$$
sends $1$ to $x=y^2$.
So, from this we see that the sheaf condition dictates that the map
$$(\Z/3\Z)^2=\underline{\Z/3\Z}(V_1)\times\underline{\Z/3\Z}(V_2)=\underline{\Z/3\Z}(U_2)\to \mu_3(U_2)$$
is given by
$$(a,b)\mapsto x^a y^{2b}$$
unless I've made a clerical error.
TL;DR: You didn't use the full presheaf compatability condition.
Best Answer
In general, if $f : (X, \mathcal O_X) \to (Y, \mathcal O_Y)$ is an open immersion of ringed spaces with image $U$ and $\mathcal F$ is a sheaf of $\mathcal O_X$-modules, you can define the presheaf of $\mathcal O_Y$-modules by $$ f_{p!}\mathcal{F}(V) = \begin{cases} \mathcal{F}(f^{-1}V), & \text{if } V \subset U \\ 0, & \text{if } V \not\subset U. \end{cases} $$ The sheaf of $\mathcal O_Y$-modules associated is denoted by $f_!\mathcal F$. You can show that $f_!$ is a functor between the categories of sheaves of $\mathcal O_X$-modules and sheaves of $\mathcal O_Y$-modules (it is even an exact functor).
Exercise 1: The functor $f_!$ is a left-adjoint of the functor $f^*$, in other words for every sheaf of $\mathcal O_X$-modules $\mathcal F$ and for every sheaf of $\mathcal O_Y$-modules $\mathcal G$, we have a natural bijection : $$\operatorname{Hom}_{\mathcal O_X} (\mathcal F, f^*\mathcal G) \simeq \operatorname{Hom}_{\mathcal O_Y} (f_! \mathcal F, \mathcal G)$$
Exercise 2: There exists a natural bijection $$\operatorname{Hom}_{\mathcal O_X}(\mathcal O_X, \mathcal F) \simeq \mathcal F(X)$$
If we adopt the notation $j$ instead of $f$ and we denote the inclusion by $j : U \to X$, the two exercises should help you prove the isomorphism that you want.