what is the meaning of $\mathbb{R}\oplus\mathbb{R}$? what is the  meaning $\mathbb{R}\oplus\mathbb{R}$?
My  confusion is that
Is  $\mathbb{R}\oplus\mathbb{R}\cong 2\mathbb{R}$  or $\mathbb{R}\oplus\mathbb{R} \cong \mathbb{R}^2$?
 A: In general, for abelian groups (or, here, vector spaces, I guess) $X$ and $Y$, we can form a new abelian group/vector space, the direct sum $X\oplus Y$, by formally adding elements of $X$ with elements of $Y$.
So the symbol $x\oplus y$ with $x\in X$ and $y\in Y$ cannot be reduced further. It's a formal symbol. We then define $x\oplus y+x'\oplus y'=(x+x')\oplus (y+y'),$ because all of these additions are assumed to already make sense (since $X$ and $Y$ have the prerequisite algebraic structure). In the vector space case, for a scalar, $\lambda,$ we define $\lambda(x\oplus y)=(\lambda x)\oplus (\lambda y),$ since this is the rule we would get if we really could add these things.
Now, you run into confusion because, in your case, $X=Y,$ so you could already add $x$ and $y$ (just use the structure in $X$), but in the direct sum, we forget this.
So, for $x,y\in \mathbb{R}\oplus \mathbb{R},$ $x\oplus y\neq x+y$. However, as a vector space, it is clearly spanned by $1\oplus 0$ and $0\oplus 1$, and hence, $\mathbb{R}\oplus \mathbb{R}$ is isomorphic to $\mathbb{R}^2$.
A: If you have a vector space $V$ and two subspaces $W_1,W_2\subseteq V$, then one defines
$$W_1+W_2=\{w_1+w_2\mid w_1\in W_1,w_2\in W_2\}.$$
In this context, $W_1\oplus W_2$ means the exact same thing as $W_1+W_2$, but should emphasize that the chosen subspaces $W_1$ and $W_2$ have trivial intersection, that means, $W_1\cap W_2=\{0\}$. This has all kinds of nice consequences, e.g. that for every $w\in W_1\oplus W_2$ there is exactly one way to write $w=w_1+w_2$ with $w_1\in W_1$ and $w_2\in W_2$.
Now, if your vector spaces $W_1$ and $W_2$ are not given as subspaces of a common space $V$, then $\oplus$ is defined to resemble the behavior that I explained above, but in an abstract way.
In your specific example, this means the following. In $\Bbb R\oplus \Bbb R$ you will enforce that these two $\Bbb R$'s (despite they look the same) are considered as distinct subspace of a bigger (but not further specified) vector space with trivial intersection. E.g., we can imagine that both of them are subspaces of $\Bbb R^2$, the first one as $\Bbb R\times\{0\}\subseteq\Bbb R^2$, and the second one as $\{0\}\times\Bbb R\subseteq\Bbb R^2$. You then can apply above definition of $\oplus$ for subspaces, that is
$$\Bbb R\oplus\Bbb R\cong \{(x_1,0)+(0,x_2)\mid x_1\in\Bbb R,x_2\in\Bbb R\}=\{(x_1,x_2)\mid x_1\in\Bbb R,x_2\in\Bbb R\}=\Bbb R^2.$$
