Here's my understanding of the intuition behind the Hilbert metric's utility for Perron-Frobenius. (I don't have access to Birhoff's paper handy so I'm not sure to what extent I'm just duplicating what's there.)
(1) As Suvrit's comment pointed out, since studying eigenvectors of $A$ is really a projective question, it is completely natural to consider a metric on the projectivization of $\mathbb{R}_+^n$, so that we are considering lines $\ell$ through the origin into the positive orthant.
(2) For our metric to be useful, we should be able to compute $d(\ell_1,\ell_2)$ in terms of $x_1,x_2$ for some (any) $x_i\in \ell_i$. This should be independent of the choice of $x_i$; replacing $x_i$ with another point on $\ell_i$ should not change the value of $d$. At the risk of being a little bit vague, this replacement is a kind of projective transformation, and so it is natural to ask our metric to be associated to a projective invariant.
(3) The most fundamental projective invariant is the cross-ratio. Given four collinear points $x_1,x_2,x_3,x_4$, the cross-ratio is
$$
(x_1,x_2;x_3,x_4) = \frac{|x_3 - x_1|}{|x_3-x_2|} \cdot \frac{|x_4-x_2|}{|x_4-x_1|}\qquad\qquad (*)
$$
(4) A cross-ratio requires four points but a metric only has two points as input. So given $x_1,x_2\in \mathbb{R}_+^n$, we need to choose two other points that are collinear with $x_1,x_2$. Let $L$ be the line through $x_1,x_2$; the only other natural reference points to choose on $L$ are the points where it intersects the boundary of the positive orthant; that is, the two points $x_3,x_4\in L$ where one (or more) of the coordinates vanishes and the rest are positive.
(5) With $x_i$ as above, note that $(x_1,x_2; x_3,x_4)$ is equal to 1 iff $x_1=x_2$ (since we always have $x_3\neq x_4$) and so to produce something with the right scaling for a metric we should put $d(x_1,x_2) = |\log(x_1,x_2;x_3,x_4)|$. This defines the Hilbert metric.
(6) It remains to get some intuition for why the Hilbert metric is a natural choice for something that is contracted by $A$. First note that since the quantities $x_i-x_j$ in $(*)$ are all scalar multiples of each other, we have $(Ax_1,Ax_2;Ax_3,Ax_4) = (x_1,x_2;x_3,x_4)$. Let $x_3',x_4'$ be the boundary points for the line through $Ax_1,Ax_2$, so that we have the following picture; note that this is where we use positivity of $A$ to guarantee that boundary points of the positive orthant ($x_3,x_4$) are mapped into the interior.
Now we have $d(x_1,x_2) = |\log(Ax_1,Ax_2;Ax_3,Ax_4)|$ and $d(Ax_1,Ax_2) = |\log(Ax_1,Ax_2;x_3',x_4')|$. The first cross-ratio is the product of the ratios
$$\frac{|Ax_1-Ax_3|}{|Ax_2-Ax_3|} \text{ and } \frac{|Ax_4 - Ax_2|}{|Ax_4 - Ax_1|},$$
while the second is the product of the ratios
$$\frac{|Ax_1-x_3'|}{|Ax_2-x_3'|} \text{ and }\frac{|x_4' - Ax_2|}{|x_4' - Ax_1|}.$$
Compare the first members of these pairs of ratios; to go from one to the other we add $|Ax_3 - x_3'|$ to both the numberator and denominator, which has the effect of making the ratio closer to the value 1. Similarly for the second ratio in each pair. Thus the cross-ratio involved in the definition of $d(Ax_1,Ax_2)$ is closer to 1 than the cross-ratio involved in $d(x_1,x_2)$, which is equivalent to the statement that $A$ contracts the metric $d$.
Of course one has to be a little more careful with this to guarantee that the contraction is uniform (and I've been a little glib regarding the relative orders of $x_1,x_2,x_3,x_4$), but this is the geometric intuition behind the fact that positive matrices contract the Hilbert metric on the positive orthant.