In this answer we make no initial assumptions about the signs of $a,b,c,d$ and go for a characterization of when the desired inequality holds or does not hold.
Let $P=c/d-a/b=(bc-ad)/(bd),$ so that $P>0$ is the condition that $a/b<c/d$ which we are assuming. [Note that $bc-ad \neq 0$ since we are assuming $a/b<c/d.$] Also put $m=(a+c)/(b+d)$ which is the "middle term" of the desired inequality $a/b<m<c/d.$ Our claim is that this inequality holds if and only if $b,d$ have the same sign.
Now define
$$\Delta_1=m-a/b=(bc-ad)/(b(b+d))=P\cdot (d/(b+d)),\\
\Delta_2=c/d-m=(bc-ad)/(d(b+d))=P \cdot (b/(b+d)).$$
Since $P>0,$ the desired inequality is thus equivalent to saying that
$$\frac{d}{b+d}>0,\ \ \frac{b}{b+d}>0.\tag{1}$$
We are of course assuming none of $b,d,b+d$ are zero, in order that the three terms entering into the inequality all be defined.
Now first suppose $b,d$ have opposite signs. Then whatever the sign of $b+d$ be, one sees from (1) that the fractions mentioned in (1) cannot each be positive, so that in this case the desired inequality does not hold.
On the other hand, supposing $b,d$ have the same signs, it follows that both fractions in (1) are positive, so the desired inequality holds.