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 $i_1$ which means that $M=M_{i_1}$.

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 $i_1$

Addendum: 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 $M=M_{i}$.