It suffices to show that if $K_i\to M_i\to N_i$ is exact at $M_i$ for each $i$ (and the appropriate squares commute), then $\varinjlim K_i\to \varinjlim M_i\to \varinjlim N_i$ is exact at $\varinjlim M_i$. 

(To get that $\varinjlim$ sends short exact sequences to short exact sequences from this, simply apply the argument to $0\to K_i\to M_i$, exact at $K_i$; to $K_i\to M_i\to N_i$; and then to $M_i\to N_i\to 0$). 

Call the first map $f_i$ (with induced map of limits $f$), the second $g_i$ (with induced map $g$); use $\kappa$, $\mu$, and $\nu$ for the structure maps. 

Given $[(k,i)]$, we have $g(f([(k,i)])) = g([f_i(k),i]) = [g_i(f_i(k)),i] = [0,i]$, so the composition is trivial. That is, $\mathrm{Im}(f)\subseteq \mathrm{Ker}(g)$.

Now assume that $g([(m,i)]) = [(0,j)]$. Then there exists $t\geq i$ such that $\nu_{it}(g_i(m)) = 0$; hence $g_t(\mu_{it}(m)) = 0$, so by exactness of the original diagram we know that there exists $k\in K_t$ such that $f_t(k) = \mu_{it}(m)$. Therefore,
$$f([k,t]) = [(f(k),t)] = [(\mu_{it}(m),t)] = [(m,i)],$$ so $[(m,i)]$ lies in the image of $f$. Thus, $\mathrm{Im}(f)\supseteq \mathrm{Ker}(g)$, proving equality.

Now, this proves that $\varinjlim$ is exact; as to "why" it is exact (is there some deep why it is exact)? I don't know if I can answer.

Why do direct limits preserve exactness?
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