Let me first note that what Kashiwara and Schapira call "left exact" is a considerably stronger condition than what most people think of, namely, the preservation of finite limits. The Yoneda embedding is always left exact in this weaker sense – in fact, it preserves all limits than exist in $\mathcal{C}$. (This is an easy exercise and amounts to unwinding definitions.) We will need to make use of this fact.
For clarity, let me call a functor that is left exact in the strong sense "representably flat".
Theorem. Suppose all idempotents in $\mathcal{C}$ split. Then, the Yoneda embedding is representably flat if and only if $\mathcal{C}$ has all finite limits.
Proof. First, assume $\mathcal{C}$ has all finite limits. Let $P$ be any presheaf on $\mathcal{C}$. Recalling that a category with all finite limits is automatically cofiltered, to show that $(P \downarrow H_\bullet)$ is cofiltered, it is enough to show that it has all finite limits. But $H_\bullet$ preserves finite limits and the forgetful functor $(P \downarrow H_\bullet) \to \hat{\mathcal{C}}$ creates them, thus, $(P \downarrow H_\bullet)$ is indeed cofiltered. (This can also be shown by hand using elementary methods.)
Conversely, suppose the Yoneda embedding is representably flat. To show that $\mathcal{C}$ has a terminal object, consider the cofiltered category $(1 \downarrow H_\bullet)$. It is non-empty, so there is a morphism $1 \to H_c$. By unwinding definitions, this means we have a morphism $f_d : d \to c$ for each object $d$ in $\mathcal{C}$ such that $f_d \circ k = f_{d'}$ for all morphisms $k : d' \to d$ in $\mathcal{C}$. In particular, $f_c : c \to c$ must be idempotent and splits as $f_c = s \circ r$ for some $r : c \to e$, $s : e \to c$ such that $r \circ s = \textrm{id}_e$. Notice that we have a morphism $d \to e$ for any $d$, namely $r \circ f_d$. But for any other $g : d \to e$, we must have
$$f_d = f_c \circ s \circ g = s \circ r \circ s \circ g = s \circ g$$
and therefore $r \circ f_d = g$ for all $g : d \to e$; hence $e$ is a terminal object of $\mathcal{C}$.
Now, let $x$ and $y$ be any two objects of $\mathcal{C}$. To show that $x \times y$ exists in $\mathcal{C}$, we consider the cofiltered category $(H_x \times H_y \downarrow H_\bullet)$. We already know this is a non-empty category because we have the two projections $\pi_1 : H_x \times H_y \to H_x$ and $\pi_2 : H_x \times H_y \to H_y$, but since it is cofiltered, we also get a morphism $f : H_x \times H_y \to H_c$ and morphisms $p_1 : c \to x$, $p_2 : c \to y$ such that $H_{p_1} \circ f = \pi_1$ and $H_{p_2} \circ f = \pi_2$. Unwinding definitions, this means for every pair $g : d \to x$, $h : d \to y$, there is a morphism $f(g, h) : d \to c$ such that $p_1 \circ f(g, h) = g$ and $p_2 \circ f(g, h) = h$. Moreover, for any $k : d' \to d$, we have $f(g, h) \circ k = f(g \circ k, h \circ k)$. In particular,
$$f(p_1, p_2) \circ f(p_1, p_2) = f(p_1 \circ f(p_1, p_2), p_2 \circ f(p_1, p_2)) = f(p_1, p_2)$$
so $f(p_1, p_2) : c \to c$ is idempotent. Suppose $f(p_1, p_2) = s \circ r$ is a splitting, where $r : c \to e$ satisfies $r \circ s = \textrm{id}_e$. I claim $e$ is the product of $x$ and $y$ in $\mathcal{C}$, with projections given by $p_1 \circ s$ and $p_2 \circ s$. Indeed, suppose $\ell : d \to e$ is any morphism such that $p_1 \circ s \circ \ell = g$ and $p_2 \circ s \circ \ell = h$. Then,
$$r \circ f(g, h) = r \circ f(p_1 \circ s \circ \ell, p_2 \circ s \circ \ell) = r \circ f(p_1, p_2) \circ s \circ \ell = r \circ s \circ r \circ s \circ \ell = \ell$$
as required for a product.
Finally, let $g, h : x \to y$ be any two morphisms of $\mathcal{C}$. To show that the equaliser of $g$ and $g$ exists in $\mathcal{C}$, we consider the cofiltered category $(E \downarrow H_\bullet)$, where $E$ is the equaliser of $H_g, H_h : H_x \to H_y$ in $\mathcal{C}$. Since the category is cofiltered, there exists a morphism $f : E \to H_c$ and a morphism $i : c \to x$ such that $H_i \circ f$ is the canonical inclusion $E \to H_x$ and $g \circ i = h \circ i$. Unwinding definitions, this means for any two morphisms $j : d \to x$ such that $g \circ j = h \circ j$, there exists a morphism $f(j) : d \to c$ such that $i \circ f(j) = j$, and for any morphism $k : d' \to d$, we have $f(j \circ k) = f(j) \circ k$. Therefore,
$$f(i) \circ f(i) = f(i \circ f(i)) = f(i)$$
and we can split $f(i)$ as $s \circ r$ for some split epimorphism $r : c \to e$. By this point it should be clear that $e$ is the equaliser of $g$ and $h$ in $\mathcal{C}$. Let us check that it works. Given any $\ell : d \to e$ such that $i \circ s \circ \ell = j$, we must have
$$r \circ f(j) = r \circ f(i \circ s \circ \ell) = r \circ f(i) \circ s \circ \ell = \ell$$
and so $e$ is indeed the equaliser of $g$ and $h$, with canonical inclusion $i \circ s$.　◼
According to §5, $\Lambda$ is the functor taking a presheaf $P$ on $X$ and returning the corresponding étale space $\Lambda_P$, which is a topological space with a natural map $p:\Lambda_P\to X$, which is a local homeomorphism.
Explicitly, as a set $\Lambda_P$ is the disjoint union $\coprod_{x\in X}P_x$ of the germs $P_x$ at each point $x\in X$, and the map $p$ just sends all elements in $P_x$ to $x$. As a topological space, well you should read the corresponding section of §5, it is explained rather well I think, but basically it is the topology such that the continuous sections $s:U\to \Lambda_P$ of $p$ are exactly the sections of the sheafification of $P$.
Now you say you have "little experience" with germs, but I don't know how you are expecting to solve exercises from a book on sheaves without diving into those notions.
Best Answer
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.