If $V$ is finite dimensional, then you can either use i) or ii) for the definition of positive definiteness, although it seems i) is stronger. To see this, apply ii) to the unit ball in $V$, which is compact, hence there is $\delta>0$ such that \begin{equation}
<Ax,x>\ge\delta
\end{equation} for all $x$ on the unit ball, then linearity leads to i). Therefore the two definitions are equivalent for finite dimensional cases.
But i) is indeed stronger than ii) in infinite dimensional spaces.
According to my limited knowledge, the usefulness of positive definiteness lies in the fact that the bilinear form\begin{equation}
(x,y)\mapsto <Ax,y>
\end{equation} is bounded from below uniformly, as in i). (Of course it is bounded from above if you assume $A$ is a bounded linear operator.)
This boundedness give very nice results, see the theorem by Lax-Milman, which is not true if you only assume ii), which allows the evil existence like \begin{equation}
<Ax_n,x_n>\to 0.
\end{equation}
One illustrating example comes from the existence of weak solutions to second order elliptic PDEs, which are of the form \begin{equation}
Lu=-\sum_{i,j}\partial_j (a_{i,j}\partial_i u)+\sum_i \partial_i(b u)+cu,
\end{equation} where the coefficients $a_{i,j}$, $b$ and $c$ are smooth functions.
Here we have the notion of ellipticity, which says $(a_{i,j}(x))$ (which is a finite dimensional matrix at each $x$) is positive definite, and uniform ellipticity, which says there is $M>0$ such that $<a_{i,j}(x)y,y>\ge M|y|^2$ for all $x$. The latter notion guarantees the existence of weak solutions. SO again you see bounded from below uniformly is important. (Here uniform in $x$).
Best Answer
Restricting to the unit ball is always illustrating. Let $A$ be a given positive definite matrix, then there is $\delta>0$ such that \begin{equation} <Ax,x>\ge\delta \end{equation} for all $\|x\|=1$.
We use the 2-norm, defined by \begin{equation} \|A\|=\operatorname{sup}_{\|x\|=1}\|Ax\|, \end{equation} which is equivalent to any other norms.
If $B$ is very close to $A$, say, $\|B-A\|<\epsilon$, then \begin{equation} |<Bx,x>-<Ax,x>|=|<(B-A)x,x>|<\epsilon\|x\|^2, \end{equation} so if you restrict to the unit ball again then you can bound $<Bx,x>$ from below using positive definiteness of $A$ and controlling $\epsilon$, and this will lead to the positive definiteness of $B$.