$\newcommand{\i}{\mathrm{i}}\newcommand{\slash}[1]{\!\not{\!#1}\,}\newcommand{\ex}[1]{\mathrm{e}^{#1}}$The title of the book that you are reading suggests that you want to search for anomalies. The typical procedure to check whether a continuous global symmetry is anomalous or not is the following. A continuous global symmetry implies the existence of a conserved current, $J_\mu$, such that $\partial^\mu J_\mu = 0$. Then, if $S_0$ is the original action of your theory, you couple the theory to a background gauge field, minimally. That is a gauge field that you are not integrating over in the path integral. Minimally means that you add to your action the term
$$\int \mathrm{d}^d x\ A_\mu J^\mu,$$
where $A_\mu$ is the gauge field. Now the symmetry presumably holds in a local manner and the gauge field transforms under it. Let's say that under the symmetry $A\mapsto A^{(\lambda)}$ where $\lambda$ is the (background) gauge parameter. The question is if the symmetry is still there at the quantum level1, i.e. if
$$Z[A^{(\lambda)}]\overset{?}{=}Z[A],\tag{1}$$
where
$$Z[A]:= \int \mathrm{D}\Phi \exp\!\Big(\underset{\i S_1[\Phi,A]}{\underbrace{\i S_0[\Phi] + \i \int \mathrm{d}^d x \ A_\mu J^\mu}}\Big),$$
where I collectively denoted all the fields $S_0$ comprises of as $\Phi$. If the answer to (1) is positive, the symmetry is not-anomalous. If it's negative, it is anomalous. This is the name of the game.
In the example you are confused about, $S_0$ is the bare Dirac action
$$S_0 = \int\mathrm{d}^4x \ \bar{\psi}\,\mathrm{i}\slash{\partial}\psi.$$
It has two global symmetries that we are interested in. One is a normal $\mathrm{U}(1)$ symmetry, under which $\psi\mapsto \ex{\i \alpha}\psi$, with $\alpha\in\mathbb{R}/2\pi\mathbb{Z}$. This one is dubbed "vector", because both the top and the bottom components of $\psi$ transform the same way. The other one, is again a $\mathrm{U}(1)$, this time transforming the top and the bottom components of the spinor with the opposite sign. This one is called "axial". Under this, $\psi\mapsto \ex{\i \beta \gamma_5}\psi$, where $\beta\in\mathbb{R}/2\pi\mathbb{Z}$. The conserved currents associated with these symmetries are $$J^\text{(vector)}_\mu = \bar{\psi} \gamma_\mu \psi \qquad\text{and}\qquad J^\text{(axial)}_\mu = \bar{\psi} \gamma_\mu \gamma_5\psi.$$ Therefore, according to the above procedure, you want to couple them to background gauge fields to check if they're anomalous. You can see then that
$$A^\text{(vector)}_\mu J_\text{(vector)}^\mu = \bar{\psi}\slash{\!A}^\text{(vector)}\psi $$
and similarly
$$A^\text{(axial)}_\mu J_\text{(axial)}^\mu = \bar{\psi}\slash{\!A}^\text{(axial)}\gamma_5\psi,$$
This answers what the vector and axial gauge fields are; the gauge fields that couple to the vector and axial conserved currents and transform accordingly, respectively. It is then clear that your (4.11) Lagrangian is the Lagrangian for the action
$$S_1[\psi,\bar\psi,A^\text{(vector)},A^\text{(axial)}] = \int\mathrm{d}^d x\ \bar\psi\left(\i\slash{\partial}+\slash{\!A}^\text{(vector)} +\slash{\!A}^\text{(axial)}\gamma_5\right)\psi,$$
as prescribed in my first paragraph. This answers why we consider the above action (equivalently your (4.11)).
From here you go on (and presumably that's what the book does) to write down
$$ Z[A^\text{(vector)},A^\text{(axial)}] = \int\mathrm{D}\bar\psi\;\mathrm{D}\psi\ \ex{\i S_1[\psi,\bar\psi,A^\text{(vector)},A^\text{(axial)}]} $$
and find whether any of the symmetries is anomalous. I won't spoil the answer.
1 Sometimes anomalies are even apparent at the classical level, see related phys.SE answer
