EDITs: 1) edited to make it work for the general case of a short exact sequence 2) edited some steps following Torsten's comments below.
I believe the following definition is equivalent to Martin's: Let $M$ be an object of an ableian category. $M$ is of finite type if for any directed system of objects $\{P_i\}$ admitting maps $\{P_i\to M\}$ consistent with the directed system $\{P_i\}$ such that the induced $\lim P_i\to M$ is surjective, there exists a $j$ such that $P_i\to M$ is surjective for all $i\geq j$.
Proposition.
Let $0\rightarrow M'\rightarrow M\rightarrow M''\rightarrow0$ be exact with $M'$ and $M''$ of finite type. Then $M$ is of finite type.
Proof: Let $(M\to Q_i):={\rm coker} (P_i\to M)$ so one has exact sequences: $$ P_i \to M \to Q_i \to 0. $$ Then $$ \lim P_i \to M \text{ is surjective} \quad \Leftrightarrow \quad \lim Q_i =0. $$ Now let $(K'_i\to M'):=\ker (M'\to M\to Q_i)$ and $(Q_i\to C'_i):={\rm coker } (M'\to M\to Q_i)$. By construction we have surjective maps: $$ M \to Q_i \to C_i $$ that composed with $M'\to M$ is the zero map. Hence we obtain a surjective map $\gamma_i : M''\to C'_i$. Similarly, for the surjective map $M \to \lim Q_i \to \lim C_i$ composed with $M'\to M$ we obtain an induced map $\gamma : M\to \lim C_i$. Observe that $\gamma$ has to be the same as $\lim\gamma_i$. However, since $\lim Q_i =0$, it follows that $\gamma=0$ and hence $\lim \gamma_i=0$. Let $K_i'':=\ker\gamma_i$. Then it follows that $\lim K_i''\to M''$ is surjective and hence for some $j$, $K''_i\to M''$ is surjective for all $i\geq j$. However, that implies that $\gamma_i=0$ and hence $C_i=0$ for $i\geq j$.
This in turn implies that $M'\to Q_i$ is surjective for $i\geq j$. Since $\lim Q_i =0$, it then follows that $\lim K'_i\to M'$ is surjective and hence there exists a $j'\geq j$ such that $K'_i\to M'$ is surjective for all $i\geq j'$. It follows that (the surjective) $M'\to Q_i$ is the zero map. Therefore $Q_i=0$, and hence $P_i\to M$ is surjective for $i\geq j'$. Q.E.D.