At least if we have a Grothendieck category everything seems OK: Suppose
$0\rightarrow M'\rightarrow M\rightarrow M''\rightarrow0$ is exact with $M'$ and
$M''$ of finite type. Assume $\{M_i\}$ is a directed collection of subobjects of
$M$ such that $\sum_i M_i=M$. We then have $M'=M'\bigcap\sum_i M_i=\sum_i
M'\bigcap M_i$ and hence $M'=M'\bigcap M_{i_0}$ for some $i_0$. Throwing away
all indices which are not $\geq i_0$ we may assume $M'\subseteq M_i$ for all
$i$. We then get that $M''=\sum_i M_i/M'$ and hence $M''=M_{i_1}/M'$ for some
$i_1$

<b>Addendum</b>: Stealing some ideas from Sándor's reply we can get the
statement without extra axioms. Note that  finite generation is formulated in
terms of $\sum_iM_i=M$ which is the same as $\mathrm{lim}M_i\to M$ being surjective
(as the sum is image of the limit). Now, with notations as before we put $M''_i$
to be the image of $M_i$ in $M''$. As $\mathrm{lim}M_i\to M$ is surjective we get
that so is $\mathrm{lim}M_i''\to M''$ and hence $M''_i=M''$ for some $i$ and after
throwing away we can assume this is always true. This means that we get an exact
sequence $0\to M'_i\to M'\to M/M_i\to0$ and as $\mathrm{lim}M/M_i=0$ (by right exactness of
directed colimits) we get that $\mathrm{lim}M'_i\to M'$ is surjective (again by
right exactness) and hence that $M'_i=M'$ for some $i$ but then $M_i=M$ as $M''_i=M''$,
 which means that $M=M_{i}$.