Let X be a topological vector space,we know bounded set $B$ defined as for any neiborhood of origin,there exist some $\lambda$ such that $B\subset \lambda U$.
Does this depend on the base point we choose (for example here we choose definition for the neiborhood at origin.) Can we show this definition is equivalent to given point $x\ne 0 $ any neiborhood $U_x$ exist $\lambda$ such that $V\subset \lambda U$?
No since bounded set under translation(which is continuous linear map) is also bounded
If you choose $x=(1,0)$ in $\mathbb R^2$, then the unit ball $B=B_1(0,0)$ will not be bounded according to your definition. This is because if $U=B_{1/2}(x)$, then one would never have $$ B\subseteq\lambda U, $$ because $(0,0)$ is not in $\lambda U$, unless $\lambda=0$, which clearly doesn't work either.
As an alternative, instead of $\lambda U$, one may consider homothety relative to $x$, meaning the map $$ h_\lambda(u) = \lambda(u-x) + x = \lambda u + (1-\lambda)x. $$
This is like multiplying by $\lambda$ when the origin is changed to $x$. It is now easy to see that a set $B$ is bounded if and only if, for any neighborhood $U$ of $x$, there exists some $\lambda$ such that $B\subseteq h_\lambda(U)$.