In topological vector spaces, translation and dilation are homeomorphism. From this one can proof that for any $x\in X$, where $X$ is the topological vector space, the neighbourhood filter $\mathcal{U}(x)$ of the point $x$ is a translation of the neighbourhood filter of the origin. Therefore, if we have the neighbourhood filter of the origin, we have the neighbourhood filters of all points in $X$. Now all authors conclude that this means that if we are given the neighbourhood filter of the origin, this completely determins the topology of that TVS. To me this is not obvious.
1.) So is there some theorem that says that if we are given the neighbourhood filters of all points of a topological (vector) space, that we can obtain the toplogy from that?
2.) In a topological vector space the topology must be such that the addition and scalar multiplication are continuous. Will the topology that we obtain from these neighbourhood filters automatically fullfill this property in a TVS?
These are two largely independent questions.
If you know that a filter is the neighborhood filter at $0$ of the topology of a TVS, then all neighborhood filters are translations of the neighborhood filter at zero, and the only question is how one can define open sets in terms of neighborhoods. This is easy. A set in a topological space is open if and only if it is a neighborhood of all of its points.
Suppose you have a filter of sets containing $0$ for a nontrivial vector space. Is it it necessarily the case that all the translates of this filter give you a topology under which the vector operations are continuous? The answer is in general no. As a counterexample, take the filter of all sets containing $0$. If you take all translates, you actually describe a topology on the underlying vector space, the discrete topology. But scalar multiplication is not continuous: If $v$ is a nonzero vector, then the sequence $\langle 1/n~ v\rangle$ does not converge to $0$.