I have some new ideas regarding the second part of your question after those three days:)
Intuition, definition of convergence
Note that by bounded we mean 'small' in some sense or 'close' to zero (if we have zero). In topology, open (nonempty ;-) ) sets are usually considered big (contrary to sets with empty interior or nowhere-dense [Baire category theorem]). Let's consider convergence - convergent sequence is also 'close' to the limit. Look at the definition of convergence: it is very similar to the definition of a bounded set in a TVS. We need all the neighbourhoods there (or at least base or subbase neighbourhoods), because one neighbourhood is just too big - even if we allow deflating it (say, a sequence is convergent to zero if there is some special neighbourhood such that - no matter how much deflated - contains the sequence without a finite number of elements). I don't know about any special subset that could be used here (of course an open ball in a normed space is great, but that's not the case).
Example justifying this intuition and how to define a subset with a 'special property'
Let $F=\{f: [0,1] \to \mathbb{R} \}$. It is a vector space over $\mathbb{R}$. Let's make it a topological vector space by establishing the topology of pointwise convergence (equivalently: product topology $\prod_{i\in [0,1]}\mathbb{R}$). A subbase of such topology is $\{A_{i,U}\}_{i \in[0,1], U=U^\circ\subseteq\mathbb{R}},$ where $A_{i,U} = \{f \in F | f(i) \in U\}.$ We can see, that in this space open base sets (even base ones!) are indeed very big: we put restrictions only on a finite number of values of $f$ - on the rest of the domain it may be whatever (in particular, arbitrarily big). And what do we get by applying the definition of a bounded set? A set $K\subseteq F$ is bounded if for each $i$ and for each subbase neighbourhood $A_{i,U}$ of the zero function, we can inflate $A_{i,U}$ to contain $K$, which means, that for each $i$ $\{f(i) | f \in K\}$ has to be bounded. In other words, $K$ is bounded, if there is $g:[0,1]\to\mathbb{R}_+$ and a set with a 'special property' $B_g$ defined as $B_g=\left \{ f \in F \ \Big| \ \forall_{i \in [0,1]} \ |f(i)| \leq g(i) \right \}$ such that $K \subseteq B_g$. We can see, that the definition is very reasonable - from the topology of pointwise convergence, we get pointwise boundness.
Nonetheless, I don't know how to define 'a set with "special property"' in a general case. Actually, it seems hard even to define $B_g$ in terms of the topology on $F$ if we do not use our specific subbase $\{A_{i,U}\}_{i \in[0,1], U=U^\circ\subseteq\mathbb{R}}$ appriopriately.
One more remark about convergence
Assume that the topology on a TVS over $\mathbb{F}$ is generated by translations of balanced sets (somewhat less than local convexity). Assume moreover for clarity, that the absolute value on $\mathbb{F}$ is nontrivial (if it is trivial, then the sequence in the proposition doesn't have to be convergent to $0$).
A set $K$ is bounded iff for (any)/(each)/(one canonical) sequence $(t_n)\in \mathbb{F}^*$ st. $t_n \to 0$ and for any neighbourhood $O$ of the $0$-vector the sequence of sets $t_n K$ falls into $O$ after finitely many steps.
Equivalently: iff for (any)/(each)/(one canonical) sequence $(t_n)\in \mathbb{F}^*$ st. $t_n \to 0$ any sequence $t_n k_n$ converges to zero (where $k_n \in K$).
This shows that boundness and convergence are very close to each other and we probably can't avoid examining all the neighbourhoods of $0$.