Do all projections matrices take this form? Do all projection matrices take the form $P = A{(A^TA)}^{-1}A^T$? If so, can you help me derive it and explain it intuitively? 

$A(A^TA)^{-1}A^T$ is symmetric, but not all projection matrices are symmetric --- such as $\pmatrix{1&1\\ 0&0}$. Thus the answer is clearly no.
It is true, however, that all orthogonal (with respect to the usual inner product) projection matrices over $\mathbb R$ can be written in the form of $A(A^TA)^{-1}A^T$. By definition, if $P\in M_n(\mathbb R)$ is an orthogonal projection, then $P|_U=\operatorname{id}$ and $P|_{U^\perp}=0$ for some subspace $U\subseteq\mathbb R^n$. Let $A$ be any matrix whose columns form a basis of $U$ (any basis will do; it doesn't have to be orthonormal). Then $A^TA$ is nonsingular and $A(A^TA)^{-1}A^Tv=0$ for every $v\in U^\perp$. Also, since the columns of $A$ span $U$, every vector $u\in U$ can be written as $Ax$ for some $x\in\mathbb R^n$. Therefore
$$
\left(A(A^TA)^{-1}A^T\right)u=\left(A(A^TA)^{-1}A^T\right)(Ax)=\left(A(A^TA)^{-1}A^TA\right)x=Ax=u
$$
for every $u=Ax\in U$. Hence $P$ and $A(A^TA)^{-1}A^T$ agree everywhere on $\mathbb R^n$, i.e. $P=A(A^TA)^{-1}A^T$.
