Suppose that $a$ is less than $b$ , $c$ is less than $d$.
What is the relation between $\dfrac{a}{b}$ and $\dfrac{a+c}{b+d}$? Is $\dfrac{a}{b}$ less than, greater than or equal to $\dfrac{a+c}{b+d}$?
Suppose that $a$ is less than $b$ , $c$ is less than $d$.
What is the relation between $\dfrac{a}{b}$ and $\dfrac{a+c}{b+d}$? Is $\dfrac{a}{b}$ less than, greater than or equal to $\dfrac{a+c}{b+d}$?
Note that if $b$ and $d$ have the same sign, then $$ \frac{a}{b}-\frac{a+c}{b+d}=\frac{ad-bc}{b(b+d)} $$ and $$ \frac{a+c}{b+d}-\frac{c}{d}=\frac{ad-bc}{d(b+d)} $$ also have the same sign.
Therefore, if $b$ and $d$ have the same sign, then $\dfrac{a+c}{b+d}$ is between $\dfrac{a}{b}$ and $\dfrac{c}{d}$.
Comment: As Srivatsan points out, if $b$ and $d$ are both positive, $$ \frac{a}{b}\lesseqqgtr\frac{a+c}{b+d}\text{ if }\frac{a}{b}\lesseqqgtr\frac{c}{d} $$
One nice thing to notice is that $$ \frac{a}{b}=\frac{c}{d} \Leftrightarrow \frac{a}{b}=\frac{a+c}{b+d} $$ no matter the values of $a$, $b$, $c$ and $d$. The $(\Rightarrow)$ is because $c=xa, d=xb$ for some $x$, so $\frac{a+c}{b+d}=\frac{a+xa}{b+xb}=\frac{a(1+x)}{b(1+x)}=\frac{a}{b}$. The other direction is similar.
The above is pretty easy to remember, and with that intuition in mind it is not hard to imagine that $$ \frac{a}{b}<\frac{c}{d} \Leftrightarrow \frac{a}{b}<\frac{a+c}{b+d}<\frac{c}{d} $$ and similar results.