Question: Assume the following bi-molecular reaction is elementary as written with rate constant $k\_\mathrm{f}$:
$$\ce{A + A -> P}$$

[This review](http://acmg.seas.harvard.edu/people/faculty/djj/book/bookchap9.html), suggests to express the rate in terms of the production of product "P":
\begin{align}
\frac{\mathrm{d}[\ce{P}]}{\mathrm{d}t} &= k\_\mathrm{f}[\ce{A}]^{2},&
\text{where }
\frac{\mathrm{d}[\ce{P}]}{\mathrm{d}t} &= \left(-\frac{1}{2}\right)\frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t},
\end{align}
which gives:
$$\frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t} = -2k\_\mathrm{f}[\ce{A}]^{2}.\tag{1}$$

While this [other review](https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/Reaction_Rates/Second-Order_Reactions) suggests that the rate be written in terms of the consumption of reactant "$\ce{A}$":
$$\frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t} = -k\_\mathrm{f}[\ce{A}]^{2}.\tag{2}$$

Which is correct?

Answer: 
If the reaction is
$$\ce{2A -> P},$$
then the rates are defined to be related as
$$
 -\frac{1}{2}\frac{\mathrm{d}\ce{A}}{\mathrm{d}t} 
 = \frac{\mathrm{d}\ce{P}}{\mathrm{d}t}
$$
and this is true throughout the reaction. The rate at which $\ce{P}$ is produced is half that at which $\ce{A}$ is consumed. Normally therefore a factor of 2 is expected with the rate constant. With any other usage an explanation should be added so as not to confuse things.

In the general case
$$\ce{aA + bB -> pP + qQ},$$
then
$$ 
 -\frac{1}{a}\frac{\mathrm{d}\ce{A}}{\mathrm{d}t} 
 = -\frac{1}{b}\frac{\mathrm{d}\ce{B}}{\mathrm{d}t} 
 = \frac{1}{p}\frac{\mathrm{d}\ce{P}}{\mathrm{d}t} 
 = \frac{1}{q}\frac{\mathrm{d}\ce{Q}}{\mathrm{d}t}
 = \frac{\mathrm{d}x}{\mathrm{d}t},
$$
where at any time $t$, $ax$ moles of $\ce{A}$ and $bx$ of $\ce{B}$ are consumed and $px$ of $\ce{P}$ are produced, where $x$ is the extent of reaction.
The first expression
$$\frac{\mathrm{d}[A]}{\mathrm{d}t} = -2k\_\mathrm{f}[\ce{A}]^{2}\tag{1}$$
is the correct one for the reaction given. As the OP mentions in the comments, both approaches describe the situation correctly. However, the values of $k\_\mathrm{f}$ would be different by a factor of two.

You could rewrite the chemical equation by dividing by two (no longer an elementary reaction, so you have to specify that it is second order in $\ce{A}$):
$$\ce{A -> 1/2 B}$$

In this case, the correct expression for the rate of $\ce{A}$ reacting would indeed be the second expression:
$$\frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t} = -k\_\mathrm{f}[\ce{A}]^{2}\tag{2}$$

Because the two reactions are written differently, you again expect a different value for $k\_\mathrm{f}$.