A function $f:R^n \rightarrow R^m$ is differentiable with respect to $\vec v$ at $\vec x_o$ if there exists a Linear transformation $L:R^n \rightarrow R^m$ such that
$$\lim_{h \to 0} \frac{f(\vec x_o + h \hat v) - f(\vec x_0) - L(h \hat v)}{h}=0$$
1) Is the above definition correct?
2) Is the linear transformation also the derivative at $x_o$ or is it just proof that the derivative exists but the derivative isn't necessarily that linear transformation?
3) If 2) is true, Is the derivative point $\vec x_o$ equal to $L(\hat v)$ or $L(\vec x_o)$?
This is a correct definition only of directional or Gateaux differentiability of $f$ at $x_0$. If $n>1$ what is usually meant by differentiability is strong or Frechet differentiability, which requires the convergence in your limit to be uniform on bounded sets of $\hat v$.
Yes: the derivative at $x_0$ is that linear transformation $L$.
I don't understand your 3d question. The derivative of $f$ at $x_0$, which is sometimes denoted $Df_{x_0}$ or $f'(x_0)$, is the linear transformation that maps $\hat v$ to $L(\hat v) = f'(x_0) \hat v$, which shows up, for instance, in the Taylor (or linearization) expansions $$f(x_0+h\hat v) = f(x_0) + h f'(x_0) \hat v + o(h)$$ (for fixed $\hat v$, for directionally differential $f$) and $$f(x_0 + v) = f(x_0) + f'(x_0) v + o(\|v\|)$$ for Frechet differentiable $f$.