The functor $[X/G]$ is not representable whenever there is non-trivial isotropy of the action of $G$ on $X$.

Let us consider the most extreme case: when $X = \bullet$ is a point (the terminal object) and $G$ is any non-trivial group. In such a case, $Hom(-,\bullet/G)$ is a singleton, as $\bullet / G = \bullet$, which is still terminal.

However, $[\bullet / G]$ is much more interesting than that! If you trace through the definition provided above (which is worth doing at least once in your life), we find that $[\bullet / G](U)$ is the collection of principle $G$-bundles over $U$, which is non-trivial most of the time. Thus it follows that, as a stack, $[\bullet / G]$ is the classifying space of $G$.

Particular examples are $[\bullet / \mathbb{G}_m]$ being the collection of line bundles, etc.

As for the question about how the morphism $[X/G] \to Hom(-,X/G)$ is defined:

In your diagram, the morphism $\alpha : E \to X$ is equivariant. You can thus complete the diagram to

$$ \begin{array}{c c c} E & \to & X \\ \downarrow & & \downarrow\\ U & \to & X/G \end{array} $$

which yields the desired map.

The functor $[X/G]$ is not representable whenever there is non-trivial isotropy of the action of $G$ on $X$.

Let us consider the most extreme case: when $X = \bullet$ is a point (the terminal object) and $G$ is any non-trivial group. In such a case, $Hom(-,\bullet/G)$ is a singleton, as $\bullet / G = \bullet$, which is still terminal.

However, $[\bullet / G]$ is much more interesting than that! If you trace through the definition provided above (which is worth doing at least once in your life), we find that $[\bullet / G](U)$ is the collection of principal $G$-bundles over $U$, which is non-trivial most of the time. Thus it follows that, as a stack, $[\bullet / G]$ is the classifying space of $G$.

Particular examples are $[\bullet / \mathbb{G}_m]$ being the collection of line bundles, etc.

As for the question about how the morphism $[X/G] \to Hom(-,X/G)$ is defined:

In your diagram, the morphism $\alpha : E \to X$ is equivariant. You can thus complete the diagram to

$$ \begin{array}{c c c} E & \to & X \\ \downarrow & & \downarrow\\ U & \to & X/G \end{array} $$

which yields the desired map.

The functor $[X/G]$ is not representable whenever there is non-trivial isotropy of the action of $G$ on $X$.

Let us consider the most extreme case: when $X = \bullet$ is a point (the terminal object) and $G$ is any non-trivial group. In such a case, $Hom(-,\bullet/G)$ is a singleton, as $\bullet / G = \bullet$, which is still terminal.

However, $[\bullet / G]$ is much more interesting than that! If you trace through the definition provided above (which is worth doing at least once in your life), we find that $[\bullet / G](U)$ is the collection of principle $G$-bundles over $U$, which is non-trivial most of the time. Thus it follows that, as a stack, $[\bullet / G]$ is the classifying space of $G$.

Particular examples are $[\bullet / \mathbb{G}_m]$ being the collection of line bundles, etc.

The functor $[X/G]$ is not representable whenever there is non-trivial isotropy of the action of $G$ on $X$.

Let us consider the most extreme case: when $X = \bullet$ is a point (the terminal object) and $G$ is any non-trivial group. In such a case, $Hom(-,\bullet/G)$ is a singleton, as $\bullet / G = \bullet$, which is still terminal.

However, $[\bullet / G]$ is much more interesting than that! If you trace through the definition provided above (which is worth doing at least once in your life), we find that $[\bullet / G](U)$ is the collection of principle $G$-bundles over $U$, which is non-trivial most of the time. Thus it follows that, as a stack, $[\bullet / G]$ is the classifying space of $G$.

Particular examples are $[\bullet / \mathbb{G}_m]$ being the collection of line bundles, etc.

As for the question about how the morphism $[X/G] \to Hom(-,X/G)$ is defined:

In your diagram, the morphism $\alpha : E \to X$ is equivariant. You can thus complete the diagram to

$$ \begin{array}{c c c} E & \to & X \\ \downarrow & & \downarrow\\ U & \to & X/G \end{array} $$

which yields the desired map.