Let V be a vector space over the field F, and W $\subset$ V a subspace. Then the linear transformation T: V $\rightarrow$ V / W is well-defined, since v $\in$ [v].
I understand v $\in$ [v], but how v $\in$ [v] implies T is well-defined? To my knowledge, being well-defined means that there is only one possible value for each T(v).
Supopse A, B and C are three vector spaces. Furthermore, assume the sequence $0\rightarrow A\rightarrow B\rightarrow C\rightarrow 0$ is exact. Then, ker($B\rightarrow C)=A$.
By the definition of the exact sequence, ker($B\rightarrow C)$=im($A\rightarrow B)$. Could you explain why ker($B\rightarrow C)=A$?
These are two separate questions.
1) $T$ implicitly sends $v$ to $[v]$. $[v]$ is unique: it describes a single member of $V/W$. Therefore $T$ is well-defined, since it sends $v$ to a unique $T(v) = [v]$.
2) Technically saying $\ker(B\to C) = A$ is a bit inaccurate, since the kernel must be part of $B$, not $A$. However, note that the short exact sequence implies that the map $A \to B$ is injective; therefore, $A$ is isomorphic to the image of $A \to B$ in $B$ (let's call this image $I$). As you noted, $I$ is the kernel of $B \to C$. Therefore, $A$ is isomorphic to the kernel of $B \to C$.