About functions of bounded variation I got the following the following idea in one of the articles that I'm reading. It goes this way. Let $X$ be a Hausdorff topological vector space and let $\mathcal{D}$ be the family of all divisions of the compact interval $[a,b]$. For function $g:[a,b]\to X$ and  each $D\in \mathcal{D}$, we write
$$V(g,D)=(D)\sum [g(v)-g(u)]$$
where
$$D=\{[u,v]\}$$
and that $u$ and $v$ are the typical division points of $[a,b]$. We say that $g$ is of bounded variation if the set
$$V_g=\{V(g,D):D\in\mathcal{D}\}$$
is a bounded subset of $X$. That is how the article introduced the concept of bounded variation in the setting of topologiacal vector spaces. 
Question: Does the concept above coincides with the standard definition of bounded variation if the space $X$ is Banach?
 A: First, let's state the definition given in the article by Weston J.D., Functions of Bounded Variation in Topological Vector Spaces, The Quarterly Journal of Mathematics, Vol 8, Issue 1, 1957, pp. 108-111:

Let $[a,b]$ be a real interval, and let $S$ be a finite succession of nonoverlapping closed subintervals $[a_i,b_i]$, $i=1,\dots, m$. Given $g:[a,b]\to X$, let 
  $$V_S(g)=\sum_{i=1}^m (g(b_i)-g(a_i)) \tag1$$ 
  and let $V(g)$ be the set of all points $V_S(g)$, for all possible choices of $S$. If $V(g)$ is bounded, $g$ is said to be of bounded variation.

The key point here is that $[a_i,b_i]$ do not have to be a partition of $[a,b]$: we can leave gaps between them. For example, if $g$ is real-valued, we would take only the intervals on which $g$ increases, or only those on which it decreases. It would not make sense to include intervals of both kinds, since it would decrease $V_S(g)$, creating cancellation in (1). 
Now, you ask about 

the standard definition of bounded variation if the space X is Banach

Well, is there the standard definition when $X$ is a Banach space? In Functional Analysis and Semi-Groups by E. Hille we find 3 definitions (see 3.2.4 on page 59):  


*

*$g:[a,b]\to X$ is of weak bounded variation if $x^*\circ g$ is of bounded variation (in the usual, real-variable sense) for every linear functional $x^*\in X^*$.

*$g$ is of bounded variation if $V(g)$, as defined above, is bounded. This is the same definition that Weston gives.

*$g$ is of strong bounded variation is $\sup_i \|g(\alpha_i)-g(\alpha_{i-1})\|<\infty$ where the supremum is over all partitions of $[a,b]$. This is probably what most people would think of as the standard definition, same as for maps into any metric space.


According to Hille, the following results are due to Dunford and Gelfand: 


*

*1 and 2 are equivalent

*3 implies 2, but not conversely


The latter is shown by an example. Take $X=L^\infty[0,1]$ and $g(t)=\chi_{[0,t]}$ for $t\in [0,1]$. Since $\|g(t)-g(s)\|_{L^\infty}=1$ whenever $t\ne s$, it's clear that $g$ is not of strong bounded variation. On the other hand, for any $0\le t<s\le 1$ the difference $g_s-g_t$ is $\chi_{[s,t]}$; summing such differences over any collection of nonoverlapping intervals we get a function of norm $1$. Thus, $g$ is of bounded variation.
