This proof maybe is too long, but anyways I decided to post it. We begin with the following definition: if $B$ is a right $R$-module, the its character module is defined by $B^*=\text{Hom}_{\Bbb{Z}}(B,\Bbb{Q}/\Bbb{Z})$. This is a left $R$-module by defining $(rf)(m)\mapsto f(mr)$ for $r\in R$ and $f\in \text{Hom}_{\Bbb Z}(B,\Bbb{Q}/\Bbb{Z})$.
We have the following lemmas:
Lemma 1: A sequence of right $R$-modules $$A_1\longrightarrow A\longrightarrow A_2$$ is exact if only if the sequence of character modules $$A_2^*\longrightarrow A^*\longrightarrow A_1^*$$ is exact.
Proof: I let you to prove this nice result.
Lemma 2: Let $R, S$ be rings. If $M$ is a finitely presented (f.p.) left $R$-module and $N$ is a $(R,S)$-bimodule, then $$\sigma:N^*\otimes_{R}M\longrightarrow \text{Hom}_{R}(M,N)^*$$ is an isomorphism.
Proof: As $M$ is f.p. then there are $m,n\in \Bbb{N}$ such that $$R^m\longrightarrow R^n\longrightarrow M\longrightarrow 0$$ is an exact sequence of left $R$-modules.
If $M=R$, then as $N^*\otimes R\cong N^*$ and $\text{Hom}_R(R,N)\cong N$ it follows that $\text{Hom}_{R}(R,N)^*\cong N^*\otimes R$. Therefore $$N^*\otimes_{R} R^m=N^*\otimes \bigoplus_{i=1}^m R \cong \bigoplus_{i=1}^m (N^*\otimes_{R} R)\cong \bigoplus_{i=1}^m \text{Hom}_R(R,N)^*=\bigoplus_{i=1}^m\text{Hom}_{\Bbb Z}\Bigl(\text{Hom}_R(R,N),\Bbb Q/\Bbb Z\Bigr)\cong \text{Hom}_{\Bbb Z}\Bigl(\bigoplus_{i=1}^m \text{Hom}_R(R,N),\Bbb Q/\Bbb Z\Bigr)\cong \text{Hom}_{\Bbb Z}(\text{Hom}_R(R^m,N),\Bbb Q/\Bbb Z)=\text{Hom}_R(R^m,N)^*.$$
Now, if we apply $N^*\otimes_R$ to the exact sequence given lines above we get the following commutative diagram
$$
\newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex}
\newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex}
\newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}}
\begin{array}{c}
N^*\otimes_R R^m & \longrightarrow & N^*\otimes_R R^n & \longrightarrow & N^*\otimes_R M & \longrightarrow & 0 \\
\da{\cong} & & \da{\cong} & & \da{\sigma} \\
\text{Hom}_R(R^m,N)^* & \longrightarrow & \text{Hom}_R(R^n,N)^* & \longrightarrow & \text{Hom}_R(M,N)^* & \longrightarrow & 0 \\
\end{array}
$$
Where the exactness of the top row follows by applying $\text{Hom}_R(\_\,,N)$ and $\text{Hom}_R(\_\,,\Bbb Q/\Bbb Z)$, noting that $\Bbb Q/\Bbb Z$ is an injective $R$-module. Finally, by the three lemma we deduce that $\sigma$ is an isomorphism.
Now we'll prove the theorem. Let
$$\begin{array}{c}
N & \ra{\phi} & N_0\longrightarrow 0
\end{array}
$$
be an exact sequence. It's enough to prove that
$$\begin{array}{c}
\text{Hom}_R(M,N) & \ra{\phi_*} & \text{Hom}_R(M,N_0)\longrightarrow 0
\end{array}
$$
is an exact sequence. Remember that $\phi_*\colon \text{Hom}_R(M,N)\rightarrow \text{Hom}_R(M,N_0)$ is defined by $\phi_*(f)=\phi\circ f$.
By lemma 1 we have that $$0\longrightarrow N_0^*\longrightarrow N^*$$ is exact. Applying $\otimes_R M$ we find the commutative diagram
$$
\newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex}
\newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex}
\newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}}
\begin{array}{c}
0 & \longrightarrow & N_0^*\otimes_R M & \longrightarrow & N^*\otimes_R M\\
& & \da{\cong} & & \da{\cong}\\
0 & \longrightarrow & \text{Hom}_R(M,N_0)^* & \longrightarrow & \text{Hom}_R(M,N)^*\\
\end{array}
$$
As $M$ is flat then the top row is exact, and the vertical maps are isomorphisms by lemma 2, so the bottom row is also exact. Thus by lemma 1 we deduce that $$\text{Hom}_R(M,N)\longrightarrow \text{Hom}_R(M,N_0)\longrightarrow 0$$ is an exact sequence. Hence, $M$ is projective.
We can use more or less the same idea to construct a counterexample in the case $\mathcal{C} = \mathbf{Set}$.
We define a functor
$$ \alpha \colon \mathbb{N}_0 \to \mathbf{Set}, n \mapsto, \{-n, \ldots, 0 , \ldots, n\}$$
where $\mathbb{N}_0$ is regarded as a filtered category by its order.
Then, $''\mathrm{colim}'' \alpha$ together with the obvious morphisms defines a group object in $\mathrm{Ind}(\mathbf{Set})$.
Now, by a lot of technical arguments one can show that this group object is not in the essential image of $\tilde{\iota_\ast}$.
Best Answer
It is not true that sequentially small modules are always finitely presented. For instance, let $A$ be an ordered abelian group with the property that every countable set of positive elements has a positive strict lower bound. Let $R$ be a valuation ring with value group $A$ and let $k$ be the residue field of $R$. Then $k$ is not finitely presented as an $R$-module, since the maximal ideal of $R$ is not finitely generated (indeed, it is not even countably generated, by our assumption on $A$). However, I claim $k$ is sequentially small.
Indeed, suppose $$M_0\to M_1\to M_2\to\dots$$ is a sequential diagram of $R$-modules with colimit $M$. Since $k$ is finitely generated, if a homomorphism $f:k\to M_n$ becomes $0$ after mapping into $B$, then it becomes $0$ after mapping into $M_m$ for some $m>n$, since we just need the images of the finitely many generators of $k$ to become $0$. So, the canonical map $\operatorname{colim}\operatorname{Hom}(k,M_n)\to\operatorname{Hom}(k,M)$ is injective.
To show it is surjective, let $f:k\to M$ be a homomorphism; we wish to show $f$ can be lifted to $M_n$ for some $n$. First, $f(1)$ can be lifted to to an element $x\in M_i$ for some $i$. Let $x_n$ be the image of $x$ in $M_n$ for each $n\geq i$, and let $I_n\subseteq R$ be the annihilator of $x_n$. If $I_n$ contains the maximal ideal for some $n$, then we can lift $f$ to a homomorphism $k\to M_n$ by mapping $1$ to $x_n$ and we're done, so we may assume $I_n$ does not contain the maximal ideal for each $n$. Then for each $n$, we can pick some $a_n>0$ in $A$ such that every element of $I_n$ has valuation at least $a_n$. By our choice of $A$, there exists $a>0$ such that $a<a_n$ for all $n$. If $r\in R$ has valuation $a$, then, the image of $ax$ in $M_n$ is nonzero for all $n$, and thus the image of $ax$ in $M$ is nonzero. But the image of $ax$ in $M$ is $af(1)=0$ since $a$ annihilates $k$. This is a contradiction.