How can I find a covariant vector in different coordinates Suppose $B$ is a covariant vector in $\mathbb{R}^{2}$ and that $B = <x^{2},x^{1}+3x^{2}>$ in the $x^{1},x^{2}$ coordinate system. Find $B$ in the new coordinate system $\bar{x}^{1},\bar{x}^{2}$ if 
$$ \bar{x}^{1} = (x^{2})^{3}$$
$$ \bar{x}^{2} = x^{1} + 2x^{2}$$
How can I do this in terms of $\bar{x}^{1},\bar{x}^{2}$?

When you say that $B$ is a covariant tensor field, you are considering the $1$-form $\omega = x^2\,dx^1 + (x^1+3x^2)dx^2$. Note that $x^2 = (\bar x^1)^{1/3}$ and $x^1=\bar x^2 - 2(\bar x^1)^{1/3}$, so
$$dx^1 = d\bar x^2 - \tfrac23(\bar x^1)^{-2/3}d\bar x^1 \quad\text{and}\quad dx^2 = \tfrac13(\bar x^1)^{-2/3}d\bar x^1.$$
Therefore 
$$\omega = (\bar x^1)^{1/3}\left(d\bar x^2 - \tfrac23(\bar x^1)^{-2/3}d\bar x^1\right) + \left(\bar x^2 - 2(\bar x^1)^{1/3}+3(\bar x^1)^{1/3}\right)\left(\tfrac13(\bar x^1)^{-2/3}d\bar x^1\right).$$
Now collect terms and find the coefficients of $d\bar x^1$ and $d\bar x^2$. These will be your new coordinates $\bar B$ of your covariant tensor field. (If I may be permitted to say so, yuck.)
If I haven't messed up, the answer is $\bar B = \langle \frac13(\bar x^1)^{-2/3}\big(\bar x^2 - (\bar x^1)^{1/3}\big), (\bar x^1)^{1/3}\rangle$.
