Since $S\cup\{v\}$ is linearly dependent and $S$ is linearly independent, then you are correct that $v$ can be written as a linear combination of vectors in $S$. That's what we're trying to show, though, so we can't use that as a fact. Instead, we'll need to use the definition of linear dependence: that there is a collection of distinct vectors $u_1,u_2,...,u_n\in S\cup\{v\}$ and a collection of non-zero scalars $a_1,a_2,...,a_n$ such that $$a_1u_1+a_2u_2+\cdots+a_nu_n=0.\tag{1}$$ Since the $a_i$s are non-zero and $S$ is linearly independent, then we can't have all the $u_i$s in $S,$ so one of them is $v$. Without loss of generality (we can always reindex if we need to), say $u_1=v$, and so the rest are in $S$. Since $a_1\neq 0,$ we can solve $(1)$ to get $$v=u_1=-\frac1{a_1}\left(a_2u_2+\cdots+a_nu_n\right)\in\text{span}\{u_2,...,u_n\}\subseteq\text{span }S.$$
Note that the definition of linear dependence I used doesn't require that we use all of the vectors in $S\cup\{v\}$ to form our dependence relation $(1)$. A set $S'$ of vectors is linearly dependent if there is some non-empty subset $\{v_1,...,v_k\}$ of $S'$ and some set of non-zero scalars $\{b_1,...,b_k\}$ such that $$b_1v_1+\cdots+b_kv_k=0.$$ We may well do this with $k=1$--for example, if the zero vector is in $S'$, we can always do this, and that's all that was meant. The set containing the zero vector could be infinite, and still be linearly dependent--the fact that the zero vector is in the set is enough for linear dependence all by itself.
Best Answer
The reverse implication is not true, consider the following three vectors in $\mathbb{R}^3$: $u_0=e_1, u_1=2e_1$ and $u_2=e_1+e_2$. The three vectors are affinely independent but not linearly independent.
The best way to understand the difference is from the picture on the wikipedia page. Think of $u_0$ as the "base vertex"and all the other $u_i$ as the positions of the other vertices. $u_i-u_0$ is then the position vector of all the other vertices relative to the base vertex, and we need these to be linearly independent (so that $u_0,\dots, u_k$ are affinely independent) so that we don't end up with three collinear vertices which would mess up our idea of what a simplex should be.
It should also be noted that the point $u_0$ is not special in the definition of affine indepenence, but in fact if $u_0,\dots,u_k$ are affinely independent then for anyfor any $j$, $u_0-u_j,\dots,u_k-u_j$ are all linearly independent. Check out this wikipedia page: http://en.wikipedia.org/wiki/Affine_space#Affine_combinations_and_affine_dependence