In the example given in Wikipedia coproduct entry, the coproduct of the family $\{X_j : j ∈ J\}$ is an object $X$ together with a collection of morphisms $I_j : X_j → X$ such that, for any object $Y$ and any collection of morphisms $f_j : X_j → Y$, there exists a unique morphism $f$ from $X$ to $Y$ such that $f_j = f ∘ I_j$.
In this specific example, I think that $\cup_{j\in J}\{X_j\}=X=\bigoplus_{j\in J}X_j$; and $X_j\oplus X_j=X_j$. Am I right?. cup is union and $\oplus$ is the coproduct.
You didn't specify a category so I will pick $\mathsf{Set}$. In $\mathsf{Set}$ a coproduct is a disjoint union. What does that mean? It means to treat the sets as disjoint when taking the union. A disjoint union of an $n$ element set and an $m$ element set is an $n + m$ element set.
So what if the sets are not disjoint? In this case, we modify the sets to make them disjoint. You may think of this as colouring the sets. So for instance we could say
$$ \{\color{red}1\} \oplus \{\color{blue}1\} = \{\color{red}1, \color{blue}1\}. $$
Now of course "modify" and "colour" are informal. We can formalize this as follows:
Modify $X_1$ and $X_2$ by means of an isomorphism (replacing them with a set of the same size).
Colour $X_1$ and $X_2$ by replacing $$X_i \leftrightarrow \{(x, i) : x \in X_i \}$$
For example $X_1 = \{0,1,2\}$ would become $\{(0,1),(1,1),(2,1)\}$ and $X_2 = \{1,2,3\}$ would become $\{(1,2), (2,2), (3,2)\}$. We have "coloured" $X_1$ and $X_2$ by putting a $1$ next to each element of $X_1$ and a $2$ next to each element of $X_2$.
Doing this we should have $$|X_1 \oplus X_2| = |X_1| + |X_2|$$ (after colouring).
It should be pointed out that $X_1 \oplus X_2$ is only determined up to isomorphism. In $\mathsf{Set}$ "isomorphic" means having the same cardinal.
Why isn't $\{1\} \oplus \{1\} = \{1\}$? (Warning: techincal details follow.) Let $X_1 = X_2 = \{1\}$. Define
$$f : X_1 \to \{1,2\} \text{ by } f(1) = 1 $$ and $$g : X_2 \to \{1,2\} \text{ by } g(1) = 2. $$
If $\{1\} = X_1 \oplus X_2$ then we are guaranteed a map $h : \{1\} \to \{1,2\}$ such that $f = h \circ \iota_1$ and $g = h \circ \iota_2$. But this is a problem because either $h(1) = 1$ which breaks $g = h \circ \iota_2$ or $h(1) = 2$ which breaks $f = h \circ \iota_1$.
As a good exercise, show that the set $\{1,2\}$ together with the maps $\iota_1 = f$ and $\iota_2 = g$ is a coproduct for $X_1$ and $X_2$. Again, coproducts are not unique but they are determined up to isomorphism.