Define the linear transformation T: P2 -> R2 by T(p) = [p(0) p(0)] Find a basis for the kernel of T. Pretty lost on how to answer this question. 
Define the linear transformation $T:P_2 \rightarrow \Bbb{R}^2$ by 
$$
T(p) =\left[\begin{array}{c}p(0)\\p(0)\end{array}\right]
$$
Find a basis for the kernel of $T$.
So a $P_2$ polynomial has the form $ax + bx + cx^2$. So $T(p)$ will always have he form $[a\; a]^\intercal$. That would mean the kernel of $T$ is $[a\; a]^\intercal$ where $a = 0$, correct? It was my understanding the kernel of a transformation is all $u$ such that $T(u)= 0$. 
If this is correct,which I'm sure it probably isn't, how do I find the basis? A basis has to be linearly independent and would have to span the kernel, so would it be a polynomial of the form $a + bx + cx^2$? 

Recall the definition of kernel. Let $V,W$ be vector spaces over the same field of scalars and let $T:V\to W$ be a linear map. The kernel of $T$ denoted by $\ker T$ is the set of all $v \in V$ such that $T(v) = 0$.
So, let $T:P_2(\Bbb R)\to \Bbb R^2$ be the map you defined, i.e.: $T(p) = (p(0),p(0))$. One arbitrary element of $P_2(\Bbb R)$ is of the form $p(x)=ax^2+bx+c$. So we have $p(0)=c$. In that case, we have the following:
$$T(ax^2+bx+c)=(c,c).$$
So, what must be $ax^2+bx+c$ to give us $T(ax^2+bx+c)=(0,0)$? It must of course be such that $c=0$. So all elements of the kernel are of the form $ax^2+bx$. And so we found:
$$\ker T=\{p \in P_2(\Bbb R) : p(x)=ax^2+bx, \quad a,b\in \Bbb R\}$$
