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 $i \circ s$. ◼
The $\text{Hom}$ functor preserves limits in each argument (in a very strong sense), neither preserves colimits in general. You should prove this as it's the source of continuity for most other things that are continuous, most notably adjoints. (Because (co)limits in functor categories are computed point-wise, this lifts to the Yoneda embeddings.)
But be a bit careful, $\text{Hom}(-,X)$ is a functor $\mathcal{C}^{op} \to \mathbf{Set}$, so being continuous means it takes colimits in $\mathcal{C}$ to limits in $\mathbf{Set}$. (Of course, this is equivalent to taking limits in $\mathcal{C}^{op}$ to limits in $\mathbf{Set}$, i.e. continuity.)
Considering a slightly different problem, if we fix an $X$ and ask what does it mean for $\text{Hom}(X,-)$ to preserve all colimits, you get the notion of a tiny object assuming $\mathcal{C}$ has all colimits (this situation is the same as the one you were talking about, just in the opposite category). An object being tiny is an unusual thing, for example, in $\mathbf{Set}$, the only tiny object is the terminal object. It would be unusual for a cocomplete category to have nothing but tiny objects. In fact, considering different classes of colimits gives different notions of "smallness". For example, if $\text{Hom}(X,-)$ preserves all filtered colimits, then $X$ is a compact or finitely presented object. A strongly finitely presented object is one for which that functor preserves all sifted colimits. A connected object (in an extensive category) is one where that functor preserves coproducts. A duality relative to a limit doctrine gives some general definitions and results for situations like this.
Best Answer
Your mistake is that they define $\mathcal C^\ast$ as the opposite category of what you have. Only then $k$ becomes covariant and only then the equality $\hom(\alpha, X)=\hom(k\alpha, k(X))$ holds. But then the colimit in the end is actually a limit in the category of sets, and both calculations will agree on the outcome.