Understanding the bra-ket antilinear correspondence 
I can't follow how the above argument leads to (1.8).
I am able to prove it only if I can show $$\langle a | c\rangle+\langle b| c\rangle=(\langle a|+\langle b|)\,|c\rangle$$
But I don't understand why the bra transformations $\langle P|$ ,$\langle Q|$ obey
$$\left(\langle P|+ \langle Q|\right)x = \langle P|x + \langle Q|x \quad .$$
Is it an assumption?
 A: First, How does the equation $1.8$ came about?
The inner product gives us a way to identify $V^*$ from $V$. If the field is complex, means there is an anti-linear correspondence between the two spaces. The vector $x\in V$ is mapped to $\langle x,\ \ \rangle $ which, since it returns a number when a vector is inserted into its vacant slot, is an element of $V^*$. This mapping is anti-linear because
$$\lambda x+\mu y\rightarrow \langle \lambda x+\mu y,\ \ \rangle =\lambda^*\langle x,\ \ \rangle +\mu^*\langle y,\ \ \rangle $$
This anti-linear map in quantum mechanics describe by saying that each vector $|\psi\rangle $ is mapped to it's dual
$$|\psi\rangle \rightarrow |\psi\rangle^\dagger \equiv \langle \psi|$$

The
$$(\langle \phi_1|+\langle \phi_2|)c=\langle \phi_1|c+\langle \phi_2|c$$
follow from the definition of vector space and $V^*$ is vector space.
A: 
Is it an assumption?

I would rather say that
$$\forall x\in V:(\langle P|+ \langle Q|)x = \langle P|x + \langle Q|x$$
is the definition of $\langle P|+\langle Q|\in V'$. More generally, given $A,B\in V'$, then $A+B\in V'$ is defined by
$$\forall x\in V:(A+ B)x = Ax + Bx.$$
