I was wondering if there is a link between this two dimension definitions in the case of a Topological Vector Space in fact I know that sometimes topological dimension coincides with other notions of dimension.
Morover are there interesting results about covering properties of TVS (such as paracompactness, refinements and so on) ?
There are various concepts of dimension for topological spcaes, e.g. covering dimension, small inductive dimension, large inductive dimension. All these take values in $\{0,1,2,\dots \} \cup \{ \infty \}$.
For algebraically finite-dimensional TVSs their algebraic dimension equals their topological dimension. This is not trivial, but well-known.
Given this result, you see that a TVS is algebraically infinite-dimensional iff it is topologically infinite-dimensional.
An algebraically infinite-dimensional TVS $E$ contains $n$-dimensional linear subspaces for all $n$, and these have topological dimension $n$. Since topological dimension is monotone (i.e. $X \subset Y$ implies $\dim X \le \dim Y$), $E$ cannot be topologically finite-dimensional.
The converse is obvious because we already know that an algebraically finite-dimensional TVS is also topologically finite-dimensional.