1. Yes, OP is right: We understand composition of operators $\hat{A}$ and $\hat{B}$ as $$(\hat{A}\circ \hat{B})(v)~:=~ \hat{A}(\hat{B}(v)), \tag{A}$$ 
where $v$ is a vector. Note that the composition symbol "$\circ$" and parenthesis "$()$" are often not written explicitly. This agrees with OP's eqs. (1) & (2). So far so good.

2. Things get surprisingly intricate, when we implement this rule (A) on the [Dirac notation](https://en.wikipedia.org/wiki/Bra%E2%80%93ket_notation). Then we define 
$$\psi(x)~:=~\langle x | \psi \rangle
~=~| x \rangle^{\dagger}| \psi \rangle, \tag{B}$$ 
$$ (\hat{A} (\psi))(x)~:=~ \langle x |\hat{A}| \psi \rangle
~=~\left(\hat{A}^{\dagger}| x \rangle\right)^{\dagger}| \psi \rangle, \tag{C} $$ 
$$ ((\hat{A}\circ \hat{B})(\psi))(x)
~:=~ \langle x |(\hat{A}\circ \hat{B})| \psi \rangle
~=~\left((\hat{B}^{\dagger}\circ \hat{A}^{\dagger})| x \rangle\right)^{\dagger}| \psi \rangle, \tag{D}$$ and so forth. Here $|x\rangle$ denotes the position ket state with eigenvalue $x$,
$$  \hat{x}|x\rangle ~=~x  |x\rangle. \tag{E}$$

3. Let us for simplicity assume that the operators $\hat{A}$, $\hat{B}$, etc, are self-adjoint. If one has never seen the lhs. of eq. (D) before, one might worry that the operators $\hat{A}$ and $\hat{B}$ seem to be composed in the wrong order! It turns out that in the end, it works out correctly after all. See e.g. the next example. 

4. _Example:_ The convention (A) implies that
$$ \hat{p}\circ \hat{x}|x\rangle ~\stackrel{(A)+(E)}{=}~x \hat{p}|x\rangle \tag{F},$$
because $\hat{p}$ is a linear operator. Therefore, we calculate
$$ ((\hat{x}\circ \hat{p})(\psi))(x)
~\stackrel{(D)}{=}~\left((\hat{p}\circ \hat{x})| x \rangle\right)^{\dagger}| \psi \rangle
~\stackrel{(F)}{=}~x\left(\hat{p}| x \rangle\right)^{\dagger}| \psi \rangle
~\stackrel{(C)}{=}~x(\hat{p}( \psi))(x),$$
as it should be. See also my related Phys.SE answer [here](https://physics.stackexchange.com/a/172089/2451).