OK, I confirm that after tensoring with $\mathbb{R}$, the map is still injective, which is actually quite easy to show.

I see! So the point is, a linear function on $\mathbb{R}^\mathbb{N}$ is basically an infinite real vector with no restriction on the coordinates. The space of linear functions is not a direct sum but a direct product. That's very helpful. Thanks a lot!

Actually I'm not quite sure why this second point is correct: if we take a (countable) real basis of $X(\bar K)\otimes_\mathbb{Z} \mathbb{R}$, the quadratic function as a sum of linear form and quadratic form, with the quadratic form determined by the induced inner product, seems that it can be determined by countably many values. Why is it uncountable?

Thanks a lot! I know very little about functional analysis; is there an easy way to think about this fact: "the space of quadratic functions on a countably-infinite-dimensional vector space has uncountable dimension"? Is it also true for linear functions?

Sorry I should have been more clear on it. I really mean the map from the infinite dimensional space to the infinite dimensional space. The quadratic function here could contain a linear part, so for example, an algebraically trivial line bundle can give a nonzero linear function. The reference I'm looking at is Lang's Fundamentals of Diophantine geometry, chapter 5.

@Will Sawin: Dear Will, I am not sure if this is correct actually. My question should have two consecutive question marks there.

Thanks for all the help! Take care!

I'm thinking about the following example: if $X=E\times C$ where $E$ is an elliptic curve and $C$ is a curve of genus $2$. Then $Sp(X)$ is supposed to be $X$. But the union of translates by closed points seems not including the codim $2$ generic point? I'm probably missing a point here. Could you help me out?

@AriyanJavanpeykar: That's very very helpful! Thanks again! Yes please make it into an answer:)

@AriyanJavanpeykar: Also, if you have a nontrivial map from a group variety to $X$, would it factor through $Sp(X)$?

@AriyanJavanpeykar: That's really helpful! Thanks a lot! So I assume if we want an argument as Proposition 3.7, we would need to know $Sp(X_L)$ is closed？Do you think there is a way to do it without assuming closedness of the set (so $Sp(X)_L$ just means the inverse image of $Sp(X)$ in $X_L$)?