Here is the construction in the case of limits.
A diagram $(X, a) : \mathsf{J} \to \int F$ consists of a diagram $X = \Pi \cdot (X, a) : \mathsf{J} \to \mathsf{C}$ along with a cone $a : * \to F\cdot X$ in $\mathsf{Set}$. Suppose that $X : \mathsf{J} \to \int F \to \mathsf{C}$ has a limit cone $\eta : \lim X \to X$ and that $F(\eta) : F(\lim X) \to F \cdot X$ is also a limit cone.
We'll show that if $\eta$ has a lift then the lift is unique. Suppose that $\eta$ lifts to a cone $\tilde{\eta} : z \to (X, a)$. Since $\Pi$ acts as the identity on arrows we must have $\tilde{\eta}_{j} = \eta_{j}$ for all $j$ in $\mathsf{J}$. So we must have $\Pi(z) = \lim X$, meaning that $z$ is of the form $(\lim X, x)$, where $x$ is an element of $F(\lim X)$. Since $\tilde{\eta} : (\lim X, x) \to (X, a)$ is a cone in $\int F$ we must have $F(\eta_{j})(x) = a_{j}$ for all $j \in \mathsf{J}$. And since $F(\eta)$ is a limit cone there is a unique element $x$ of $F(\lim X)$ satisfying these equations, so there is at most one lift of $\eta : \lim X \to X$ to a cone in $\int F$.
So now to show that $\Pi$ strictly creates limits we just have to check that this $\tilde{\eta} : (\lim X, x) \to (X, a)$ is a cone in $\int F$ and that it is limiting. I'll leave it to you to do the checking.
First of all, when the hint speaks of a "quasi-inverse" it is referring to the following equivalent of the given definition: a functor $F : \mathbf{C} \to \mathbf{D}$ is an equivalence of categories if and only if there exists a functor $G : \mathbf{D} \to \mathbf{C}$ such that $F \circ G \simeq \operatorname{id}_{\mathbf{D}}$ and $G \circ F \simeq \operatorname{id}_{\mathbf{C}}$; and in this case, $G$ is called a quasi-inverse of $F$.
So, one way to follow the hint would be to explain how $\mathbf{s}$ becomes a functor (i.e. how it operates on morphisms, and show it preserves identities and compositions), and then establish isomorphisms $\mathbf{r} \circ \mathbf{s} \simeq \operatorname{id}$ and $\mathbf{s} \circ \mathbf{r} \simeq \operatorname{id}$.
On the other hand, it is possible to proceed using the definition you stated. First, as a preliminary, I don't know if MacLane and Moerdik specified what exactly $\operatorname{Sh}(\mathcal{B})$ means; but the reasonable definition would be that it is the presheaves on the poset category of $\mathcal{B}$ such that whenever $\{ V_i \mid i \in I \} \subseteq \mathcal{B}$ is a cover of $U \in \mathcal{B}$, we have an equalizer diagram
$$F(U) \rightarrow \prod_{i\in I} F(V_i) \rightrightarrows \prod_{i, j \in I, W\in \mathcal{B}, W \subseteq V_i \cap V_j} F(W).$$
(The first step would be to see why $\mathbf{r}$ of a sheaf on $X$ would satisfy this condition; I will leave that as an exercise.)
So, first let us see that $\mathbf{r}$ is injective on morphisms; so, suppose that we have two morphisms $f, g : F \to G$ such that $f(V) = g(V)$ whenever $V \in \mathcal{B}$. Then for any open $U$ and $x \in F(U)$, there is a cover of $U$ by elements $\{ V_i \mid i \in I \} \subseteq \mathcal{B}$. Now, by the hypothesis, $$f(x) {\mid_{V_i}} = f(V_i)(x {\mid_{V_i}}) = g(V_i)(x {\mid_{V_i}}) = g(x) {\mid_{V_i}}$$ for each $i$; and by the injectivity part of the equalizer condition defining that $G$ is a sheaf, we conclude that $f(x) = g(x)$. Since this is true for any open $U$ and any $x \in F(U)$, then $f = g$.
Similarly, to see that $\mathbf{r}$ is surjective on morphisms, suppose we have $f : \mathbf{r}(F) \to \mathbf{r}(G)$. Then for any open $U \subseteq X$ and $x \in F(U)$, again choose a cover of $U$ by $\{ V_i \mid i \in I \} \subseteq \mathcal{B}$. (In fact, to forestall questions of the following construction being well-defined, let us use the canonical maximal cover of all elements of $\mathcal{B}$ contained in $U$.) Then for each $i \in I$, define $y_i := f(V_i)(x {\mid_{V_i}})$. Then for each $i,j$, we can find the canonical maximal cover of $V_i \cap V_j$ by $\{ W_k \mid k \in K_{i,j} \} \subseteq \mathcal{B}$. Now for each $k$, we have
$$y_i {\mid_{W_k}} = f(V_i)(x {\mid_{V_i}}) {\mid_{W_k}} = f(W_k)((x {\mid_{V_i}}) {\mid_{W_k}}) = F(W_k)(x {\mid_{W_k}}) = y_j {\mid_{W_k}}.$$
Therefore, by the injectivity part of the sheaf condition on $G$, we have $y_i {\mid_{U_i \cap U_j}} = y_j {\mid_{U_i \cap U_j}}$. Then, by the exactness part of the sheaf condition on $G$, there exists a unique $y \in G(U)$ such that $y {\mid_{U_i}} = y_i$. We now define $f'(U)(x) := y$.
It remains to show that $f'$ defines a morphism of sheaves, and that $\mathbf{r}(f') = f$. (Hint for the morphism of sheaves part: given $U' \subseteq U$ and $x \in F(U)$, show that $(f'(U) {\mid_{U'}}) {\mid_{V_i}}$ is equal to $y_i$ when you put $x {\mid_{U'}}$ in place of $x$, and then apply the injectivity part of the sheaf condition on $G$.)
Now, to show that $\mathbf{r}$ is essentially surjective, suppose we have $F \in \operatorname{Sh}(\mathcal{B})$. Then for each open $U$, define $G(U)$ to be the equalizer in the diagram
$$G(U) \rightarrow \prod_{V \in \mathcal{B}, V \subseteq U} F(V) \rightrightarrows \prod_{V, V', W \in \mathcal{B}, V \subseteq U, V' \subseteq U, W \subseteq V \cap V'} F(W).$$
The restriction maps of $G$ will then be constructed based on the universal property of equalizers. We now need to see that $G$ is a sheaf on $X$, and that $\mathbf{r}(G) \simeq F$. The latter follows fairly directly from the sheaf condition on $F$.
For the sheaf condition, suppose we have a cover $\{ U_i \mid i \in I \}$ of $U$ and sections $x_i \in G(U_i)$ such that $x_i {\mid_{U_i \cap U_j}} = x_j {\mid_{U_i \cap U_j}}$ for each $i,j$. Then each $x_i$ can be decomposed into the compatible data of an element of $F(V)$ for every $V \in \mathcal{B}$, $V \subseteq U_i$ which we will call $x_i {\mid_V}$. But then, the union of the canonical covers of each $U_i$ will form a cover of $U$; and for each $W$ in this cover, we can choose $i$ such that $W \subseteq U_i$, and define $y_W := x_i {\mid_W}$. If we have two different indices $i,j$ such that $W \subseteq U_i$ and $W \subseteq U_j$, then from the condition $x_i {\mid_{U_i \cap U_j}} = x_j {\mid_{U_i \cap U_j}}$ we get $x_i {\mid_W} = x_j {\mid_W}$, which makes this definition of $y_V$ well-defined. Once we verify the compatibility condition on $(y_W)$, we get a section $z_V \in F(V)$ from the definition of $F$ being a sheaf. It now remains to show that this family of $z_V$ satisfies the compatibility condition from the definition of $G$, and that the section $x \in G(U)$ we get in this way satisfies $x {\mid_{U_i}} = x_i$ for each $i$. It also remains to establish the uniqueness of $x$.
In the above, you can see that our construction in the "essential surjectivity" proof amounted to specifying the object part of a quasi-inverse $\mathbf{s}$, and our construction in the "surjectivity on morphisms" proof amounted to specifying the morphism part of $\mathbf{s}$. (Note that the definition of $\mathbf{s}$ as you wrote it does not necessarily make sense if $\mathcal{B}$ is not closed under intersections.)
Best Answer
Try to show that $G$ generally does not act continuously and consistent with the actions on $A_i$ on an infinite product in $\mathbf{Set}$ of continuous $G$-sets $A_i$. Hint: under the action induced from the $A_i$, what is the stabilizer of $(a_i)_{i\in I}$? What kind of intersections of open subgroups remain open?