lie algebras - Representations of reductive Lie group

You need to be over a field of zero characteristic and your representation needs to be rational, i.e. matrix entries need to be algebraic functions on $G$. Then it is completely reducible, see any book on algebraic groups, e.g., Jantzen or Humphreys.

You can always differentiate, so a differential of a map $Grightarrow GL(V)$ is a representation of ${mathfrak g}$. In the opposite direction, a certain care is required. To integrate a vector field, you need exponential function, which is not, in general, algebraic. However, for a semisimple group in characteristic zero, you have enough nilpotent elements $Xin{mathfrak g}$, so that the polynomials $e^{rho (X)}$ define a representation of the group.

Finally, the answer is no. Take ${mathfrak g}$ to be one-dimensional Lie algebra acting on $K^2$ by the nilpotent nonzero transformation.