Show that $\Gamma$, $\Lambda$, and the associated sheaf functor are all left exact. This is Exercise II.6 of Mac Lane and Moerdijk's, "Sheaves in Geometry and Logic [. . .]". According to the first few pages of this Approach0 search, it is new to MSE.
The Details:
The functors $\Lambda$ and $\Gamma$ are discussed in the following:

*

*Just what is Mac Lane & Moerdijk's $\Lambda$ from $\S II.5$?


*. . . and what about $\Gamma$ in $\S II.5$ of Mac Lane and Moerdijk?
Both of which are my questions.

Definition: A functor is left exact if it preserves all finite limits.

In Mac Lane's "Categories for the Working Mathematician," p. 201, we have the following.

[A functor] $T$ is left exact if and only if it is addictive and $\ker (Tf)=T(\ker f)$ for all $f$: the last condition is equivalent to the requirement that $T$ preserves short left exact sequences.

Here additive means that $T: A\to B$ is such that $A, B$ are ${\rm Ab}$-categories and
$$T(f+f')=T(f)+T(f')$$
for any parallel pair $f, f': b\to c$ in $A$, where an ${\rm Ab}$-category $C$ is a category such that each hom-set $C(p,q)$ is an additive abelian group and for which composition is bilinear: For arrows $p, p':x\to y$ and $q,q': y\to z$,
$$\begin{align}
(q+q')\circ(p+p')&=(q\circ p)+(q\circ p') \\
&+(q'\circ p)+(q'\circ p').
\end{align}$$
The Question:

Prove that the functors $\Gamma$ and $\Lambda$ and hence the associated sheaf functor
$$\Gamma\Lambda: \mathbf{Sets}^{\mathcal{O}(X)^{{\rm op}}}\to {\rm Sh}(X)$$
are each left exact.

Thoughts & Context:
The notion of a limit I am familiar with is in terms of cones and universal properties with respect to a diagram category, like on p. 21 of Mac Lane and Moerdijk. The reason I'm putting this in this section of the question (instead of The Details above) is because it's been a while since I've worked with limits, so it fits better in the context of the question.
I can (sort of) see why, if $\Gamma$ and $\Lambda$ each preserve limits, then so would $\Gamma\Lambda$.
This is the sort of question I think I could answer myself if I had more time.
I think a line of attack would be to use Proposition 3.2 of the n-lab page on exact functors:

A functor between categories with finite limits preserves finite limits if and only if:

*

*it preserves terminal objects, binary products, and equalizers; or


*it preserves terminal objects and binary pullbacks.

Please help :)

Edit: I don't think I can answer this without help.

I think I have to find the terminal object (binary products, equalisers, resp.) of the domains of each of $\Gamma$, $\Lambda$, and the associated sheaf functor then show that they each map them to that (terminal object, binary products, equalisers, resp.) of their codomains.
 A: Recall:
$\Lambda : \newcommand\O{\mathcal{O}}\newcommand\Set{\mathbf{set}}\newcommand\op{\mathrm{op}}
[\O(X)^\op,\Set] \to \newcommand\Etale{\mathbf{Etale}}\Etale(X)$ sends a presheaf to its associated etale bundle of germs.
$\Gamma : \newcommand\Bund{\mathbf{Bund}} \Bund(X) := \newcommand\Top{\mathbf{Top}}\Top/X \to \newcommand\Sh{\mathbf{Sh}}\Sh(X)$ sends an $X$-bundle to its sheaf of sections.
We'll show that they both preserve (binary) pullbacks and terminal objects.
I'm going to use $=$ when sometimes I mean natural isomorphism for notational simplicity.
Terminal objects:
The terminal presheaf is $*$, where $*(U) = \{*\}$ for all $U$.
Every stalk at any point $x$ has a unique germ, $*_x$, so $\Lambda(*)$ has underlying set isomorphic to $X$, and the open sets are the sets of the form $\{(x,*_x) : x\in U\}$, with $U$ open in $X$. Thus $\Lambda(*)$ is the trivial bundle $\mathrm{id}_X : X\to X$, which is the terminal object in $\Bund(X)$, and hence also $\Etale(X)$.
There is a unique section of the trivial bundle for any open set $U$, namely the inclusion $U\hookrightarrow X$, so $\Gamma(\newcommand\id{\operatorname{id}}\id_X) = *$, where $*$ is the terminal presheaf, but now regarded as a sheaf.
Pullbacks
Let $F,G,T$ be presheaves, with morphisms $f:F\to T \leftarrow G : g$, and let $F\times_T G$ be the pullback. Since limits commute with filtered colimits, for each $x\in X$, we have $(F\times_T G)_x = F_x\times_{T_x} G_x$, so as sets over $X$, $\Lambda(F\times_T G)$ and $\Lambda(F)\times_{\Lambda(T)}\Lambda(G)$ agree. We just need to check that their topologies are also the same. However, the open sets in $\Lambda(F\times_T G)$ are generated by the sections $s\in (F\times_T G(U))$ for all $U$, and these are the same as elements $(x,y)$ with $fx=gy$ of $F(U)\times_{T(U)} G(U)$, since limits in presheaves are computed pointwise. And these (you can check) generate the open sets in $\Lambda(F)\times_{\Lambda(T)}\Lambda(G)$. Thus the topologies are the same.
Now the pullback in $\Top/X$ is the ordinary pullback in $\Top$. Let $F : A\to X$, $G:B\to X$, and $T:C\to X$ be bundles, and $f:F\to T$ and $g:G\to T$ be maps of bundles.
A section $\Gamma(F\times_T G)(U)$ is a bundle map $\sigma :U \to F\times_T G$, where $U:U\hookrightarrow X$ is the bundle representing the inclusion of the open set $U$. 
So 
$$\begin{align}
\Gamma(F\times_T G)(U) &= \Top/X(U,F\times_T G)\\
& =\Top/X(U,F)\times_{\Top/X(U,T)}\Top/X(U,G) \\
&=\left(\Gamma(F)\times_{\Gamma(T)}\Gamma(G)\right)(U),
\end{align}$$
since limits in sheaves are still computed pointwise.
Thus $\Gamma(F\times_T G) \cong \Gamma(F)\times_{\Gamma(T)} \Gamma(G)$.
End note
You should be slightly careful and check that the morphisms defining the limit cones map correctly under the functors, but I'll leave that to you to check.
