1. Yes, OP is right: we understand composition of operators $\hat{A}$ and $\hat{B}$ as $$(\hat{A}\circ \hat{B})(v)~:=~ \hat{A}(\hat{B}(v)), \tag{1}$$ 
where $v$ is a vector. We often don't write the composition symbol "$\circ$" explicitly.

2. _Notabene:_ Note that if $|x\rangle$ denotes the position ket with eigenvalue $x$,
$$  \hat{x}|x\rangle ~=~x  |x\rangle, \tag{2}$$
then the convention (1) implies that
$$ \hat{p}\hat{x}|x\rangle ~~\stackrel{(1)+(2)}{=}~x \hat{p}|x\rangle \tag{3},$$
because $\hat{p}$ is a linear operator. Together with the [CCR](http://en.wikipedia.org/wiki/Canonical_commutation_relation) 
$$ [\hat{x},\hat{p}]~=~i\hbar{\bf 1}\tag{4}$$
this implies that
$$ \hat{x}\hat{p}|x\rangle ~\stackrel{(3)+(4)}{=}~x \hat{p}|x\rangle +i\hbar|x\rangle  \tag{5}.$$
See also [this](https://physics.stackexchange.com/q/76299/2451) Phys.SE post.