Why aren't those Cartier Divisors equivalent? Please refer to Gathmann's notes http://www.mathematik.uni-kl.de/~gathmann/class/alggeom-2002/main.pdf at Example 9.3.6 for context.
It's trying to give an example that the map between $Div(X)$ and $Z_{0}(X)$ (see text) is not injective. $X$ is a 1-dimensional scheme (a curve) given by $X=X_1\cup X_2$ where $X_i$ is isomorphic to $\mathbb{P}^1$, $i=1,2$ and they are both glued together in a point $P$ that lies in $X_1\cap X_2\neq \emptyset$. We consider the covering $U=X-\{Q\}$, $Q\in X_1-X_2$, and $V=X_1-\{P\}$.
Now consider the Cartier Divisor on X given by the constant function 1 on $U$ and the linear function that has a simple zero on $Q$ on $V\cong \mathbb{A}^1$. So we have a Cartier divisor $D$ which Weil divisor class class is $[D]=1.[Q]$.
Symetrically, we can obtain another Cartier divisor $D'$ on $X$ which Weil divisor class is given by $[D']=1.[Q']$ for a point $Q'\in X_2-X_1$.
My question is: why is the divisor $D-D'$ not linearly equivalent to zero (that is, why is it not a divisor of a rational function on $X$)?

You can use the fact that rational equivalence is preserved under pullback. Consider the inclusion map $\pi:X_2\to X$. Then you can see that $\pi_i^*D$ is trivial while $\pi^*D'$ is class of a hyperplane in $\mathbb{P}^1$. 
Edit: correction needed
This argument is technically incorrect, but I'd rather keep it in this form since when combined with Georges's comments it might explain some tricks (or even trickeries :)) in intersection theory. The correct argument should go sth like this: Let $\mathcal{D}$ and $\mathcal{D}'$ be the classes of $D$ and $D'$. Use the fact that pullback preserves linear equivalence of divisor classes (see Georges's comments on why we need to work with the class of a divisor here).  Consider the inclusion map $\pi:X_2\to X$. Then you can see that $\pi_i^*\mathcal{D}$ is trivial while $\pi^*\mathcal{D}'$ is class of a hyperplane in $\mathbb{P}^1$.
