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 $i \circ s$. ◼
First let us recall the fact that every presheaf is a colimit of representables. Just Googling this sentence should already give you plenty of proofs, but just to link two: Wikipedia and another question on math.se. That is, given a presheaf $P$ in $\mathcal{C}^*$ (keeping you notation for the category of presheaves), we have that $P = \operatorname{colim}_{i \in I} y(C_i)$ for some objects $C_i$ in $\mathcal{C}$. Here $y$ denotes the Yoneda embedding (I will use $y'$ for the Yoneda embedding $y': \mathcal{D} \to \mathcal{D}^*$, just to make clear which one is used when).
Now the proof proceeds by abstract nonsense (essentially this is working out the direction Derek tried to point towards in the comments). The proof is essentially a long chain of natural isomorphisms (or simply actual equalities). It may be a good exercise to stop at each step and see if you can do the rest by yourself. For this reason I have broken up this chain at every step, but if you wish you can just read all the centred math/text and paste it together in one long chain of isomorphisms.
Let $P$ be a presheaf in $\mathcal{C}^*$ and let $D$ be some object in $\mathcal{D}$. Then by definition
$$
\operatorname{Hom}(P, S(D)) = \operatorname{Hom}(P, F^* y'(D)).
$$
Since every presheaf is a colimit of representables we have
$$
\operatorname{Hom}(P, F^* y'(D)) \cong \operatorname{Hom}(\operatorname{colim}_{i \in I} y(C_i), F^* y'(D)).
$$
Then using that Hom-sets turn colimits in their first argument into limits we have
$$
\operatorname{Hom}(\operatorname{colim}_{i \in I} y(C_i), F^* y'(D)) \cong \lim_{i \in I} \operatorname{Hom}(y(C_i), F^* y'(D)).
$$
By the Yoneda-lemma we then find
$$
\lim_{i \in I} \operatorname{Hom}(y(C_i), F^* y'(D)) \cong \lim_{i \in I} F^* y'(D)(C_i).
$$
By definition of $F^*$ this gives us
$$
\lim_{i \in I} F^* y'(D)(C_i) = \lim_{i \in I} y'(D)(F(C_i)).
$$
By the definition of the Yoneda-embedding we then have
$$
\lim_{i \in I} y'(D)(F(C_i)) = \lim_{i \in I} \operatorname{Hom}(F(C_i), D).
$$
Once more using that Hom-sets convert limits into colimits in their first argument (here we use that $\mathcal{D}$ is cocomplete), we obtain
$$
\lim_{i \in I} \operatorname{Hom}(F(C_i), D) \cong \operatorname{Hom}(\operatorname{colim}_{i \in I} F(C_i), D).
$$
This then gives us the desired description for our left adjoint: send a presheaf $P \cong \operatorname{colim}_{i \in I} y(C_i)$ to $\operatorname{colim}_{i \in I} F(C_i)$.
Of course, to really prove that this gives you a functor, you would still need to check that this works for arrows as well. Here is an idea for what to do. In the same style as above, so you can stop at any time to try to do the rest on your own.
Let $P = \operatorname{colim}_{i \in I} y(C_i)$ and $Q = \operatorname{colim}_{j \in J} y(C_j)$. The idea is that, given an arrow $P \to Q$, to get an arrow $\operatorname{colim}_{i \in I} F(C_i) \to \operatorname{colim}_{j \in J} F(C_j)$, we will want to use the universal property of the colimit $\operatorname{colim}_{i \in I} F(C_i)$. That is, we will try to make $\operatorname{colim}_{j \in J} F(C_j)$ into a cocone for the diagram $(F(C_i))_{i \in I}$. Again, we will make some identifications, starting with
$$
\operatorname{Hom}(P, Q) = \operatorname{Hom}(\operatorname{colim}_{i \in I} y(C_i), \operatorname{colim}_{j \in J} y(C_j)).
$$
Pulling the colimit out of the first argument gives us
$$
\operatorname{Hom}(\operatorname{colim}_{i \in I} y(C_i), \operatorname{colim}_{j \in J} y(C_j)) \cong \lim_{i \in I} \operatorname{Hom}(y(C_i), \operatorname{colim}_{j \in J} y(C_j)).
$$
Then by the Yoneda lemma we get
$$
\lim_{i \in I} \operatorname{Hom}(y(C_i), \operatorname{colim}_{j \in J} y(C_j)) \cong \lim_{i \in I} (\operatorname{colim}_{j \in J} y(C_j))(C_i).
$$
Using the fact that colimits in presheaf categories are calculated pointwise we then turn this into
$$
\lim_{i \in I} (\operatorname{colim}_{j \in J} y(C_j))(C_i) = \lim_{i \in I} \operatorname{colim}_{j \in J} (y(C_j)(C_i)),
$$
which by the definition of the Yoneda embedding becomes
$$
\lim_{i \in I} \operatorname{colim}_{j \in J} (y(C_j)(C_i)) = \lim_{i \in I} \operatorname{colim}_{j \in J} \operatorname{Hom}(C_i, C_j).
$$
So to sum up, we have
$$
\operatorname{Hom}(P, Q) \cong \lim_{i \in I} \operatorname{colim}_{j \in J} \operatorname{Hom}(C_i, C_j).
$$
An arrow $P \to Q$ thus corresponds to some tuple $([f_i])_{i \in I}$ of equivalence classes of arrows in $\mathcal{C}$. If you write out the definitions here, we have that arrows $f_i: C_i \to C_j$ and $f_i': C_i \to C_{j'}$ are in the same equivalence class if they factor through the diagram $(C_j)_{j \in J}$ in essentially the same way. That is, there is $j^*$, $a: C_j \to C_{j^*}$ and $b: C_{j'} \to C_{j^*}$ such that $a f_i = b f_i'$. This means precisely that if we have a cocone of the $(C_j)_{j \in J}$, then an equivalence class $[f_i]$ determines a unique arrow into the vertex of that cocone. In particular, if we take images under $F$ we get that each equivalence class $[f_i]$ determines a unique arrow $g_i: F(C_i) \to \operatorname{colim}_{j \in J} F(C_j)$ by composing $F(f_i)$ with the coprojection of $F(C_j)$ into the colimit.
So from the tuple $([f_i])_{i \in I}$ we then obtain a tuple of arrows $(g_i)_{i \in I}$ (remember, where $g_i: F(C_i) \to \operatorname{colim}_{j \in J} F(C_j)$). Writing out the definition of the limit we obtained that tuple from, we precisely get that $(g_i)_{i \in I}$ forms a cocone for the diagram $(F(C_i))_{i \in I}$.
So we used an arrow $P \to Q$ to make $\operatorname{colim}_{j \in J} F(C_j)$ into the vertex of a cocone of $(F(C_i))_{i \in I}$. Which means that we obtain an arrow $\operatorname{colim}_{i \in I} F(C_i) \to \operatorname{colim}_{j \in J} F(C_j)$, which is the arrow we will send our original $P \to Q$ to.
Best Answer
I guess it is just the classical application of Yoneda lemma, in the form that adjoints are unique up to canonical iso: call $U$ the forgetful functor; it is just precomposition with the functor $j : Ob(J) \to J$. So, if $U$ has a left adjoint, it must be left extension along $j$.
If you now prove that the functor $L$ Emily defined is such that $$ C^{Ob(J)}(D, UX)\cong C^J(LD,X) $$ (you did it by hand, it's a reasonable way) then you get a natural isomorphism $$ C^J(LD,X) \cong C^J(Lan_jD,X) $$ and now Yoneda lemma entails that $LD\cong Lan_jD$, canonically.