Q:

Existence of incompressible surface in a non-orientable manifold.

Let $M$ be a compact $P^2$ -irreducible 3-manifold.
  If $M$ is non-orientable, then there is a compact surface $F$ properly embedded in $M$ such that $F$ is two-sided, non-separating and incompressible.
Proof: We have that $H^1(M)$ is non zero.
  In particular, there is an essential map $f$ from $M$ to the circle $S^1$. Now let $v$ be a point of $S^1$
  to which $f$ is transverse,we can perform surgery on $f^{−1} (v)$ to make it incompressible. Then $f^{−1}(v)$ is
  the required non-separating incompressible surface $F$.

What I don't understand is how do we get such an essential map and why is $f^{-1}(v)$  an incompressible surface?

A:

To solve this we need to use different machines. 

First of all, we generally have for all spaces: $H^1(X)\neq 0$ implies existence of a homotopically non-trivial $X\to S^1$, because of 

$$[X,S^1] =[X, K(\mathbb Z,1)] = H^1(X;\mathbb Z).$$

Now differential topology, more specifically transversality, will give you that the generic preimage of a $k$-submanifold of an $n$-manifold is a $(n-k)$ - codimensional submanifold. In your case that makes a surface. 
Putting the first two facts together, we see that as a generic preimage of a point of $f:M\to S^1$ representing $f_*\in Hom(\pi_1M, \mathbb Z)\cong H^1(M)$ gives a two-sided (normal bundle pulls back) properly embedded 1-codimensional submanifold $N\subset M$. $N$ defines a non-trivial 1-codimensional $\mathbb Z/2$-homology class in $M$. Hence there is a connected component $N_0\subset N$ which is homologically non-trivial in $M$. In particular $M-N_0$ is connected.
Next thing you want to do is surgery on $N_0$ --- successively surger out any compression disk you can find. If new trivial/separating components arise: throw them away (here you use the $P^2$-incompressibilty). These surgeries don't change the homology type of $N_0$.

Eventually you obtain your desired surface in $M$.
Note that 1. does not put any assumptions on the space, 2. & 3. work for general smooth manifolds, and only in 4. we actually need $dim(M)=3$.
For further readings on this (at least in the similar case with orientability; you can often either adapt the proofs for the non-orientable case or just ignore everything by substituting $\mathbb Z$ with $\mathbb Z/2$ and adding a primitivity condition) I suggest :

Codimension 1 homology represented by Embedded Submanifold
Poincare duality isomorphism problem in the book "characteristic classes"
Closed Poincaré dual of the boundary of a closed manifold.
Geometric interpretation of the evaluation of Poincaré dual with a fundamental class

