Let $A$ be an Artin algebra, $\text{mod}\,A$ the category of finitely generated $A$modules and $\text{Ab}$ the category of abelian groups. Is every additive, covariant, leftexact functor $F:\text{mod}\,A \rightarrow \text{Ab}$ (natural) isomorphic to $\text{Hom}(X,)$ for some $X\in \text{mod}\,A$? How do we obtain $X$?

4$\begingroup$ $A = k$ a countable field (e.g. $\mathbb Q$), $F(Y) = Y^{\oplus\infty}$ is a counterexample: If it was corepresented by $X$, by evaluation on $k$ we would have that $X^\vee$ is (the underlying abelian group of) a $k$vector space of countable dimension, which is impossible. Morally speaking, the corepresenting object should be a predual of $F(A)$; if you add enough assumptions on $F$ so that this exists, you should be able to prove that it is corepresenting $F$ by building $A$ via iterated squarezero extensions. $\endgroup$ – Bertram Arnold Apr 9 at 9:32

$\begingroup$ Thanks for you answer! Im my specific situation I have that $F$ is a subfunctor of $\text{Hom}(X,)$, so things like infinite sums should not occure. Do you know the answer in this case? $\endgroup$ – kevkev1695 Apr 9 at 9:37

$\begingroup$ Dan Petersen's answer is probably sufficient for your case, but being a subfunctor of a corepresentable functor, together with Artinianness, is indeed enough. I have updated my answer with a proof. $\endgroup$ – Bertram Arnold Apr 9 at 19:16
One can prove the following result: let $F \colon \mathrm{Mod}_R \to \mathrm{Mod}_S$ be leftexact and preserve small products (equivalently, a continuous functor). Then $F$ is of the form $\mathrm{Hom}(M,)$.
Let me first take a step back. The context of your question is (as I think you know) the EilenbergWatts theorem, a form of which states that if $G:{}_R\mathrm{Mod}^{\mathrm{op}} \to \mathrm{Mod}_S$ is left exact and preserves products, then there is an $(R,S)$bimodule $M$ such that $G \simeq \mathrm{Hom}(,M)$. In fact applying the functor to $R$ shows that $M =G(R)$, so the candidate module was obvious all along. And in a dual form, if $H:\mathrm{Mod}_R \to \mathrm{Mod}_S$ is right exact and preserves coproducts, then there is an $(R,S)$bimodule $M$ such that $H \simeq \otimes M$, and necessarily $M \cong H(R)$.
But in this case it's not as immediate what to plug into the functor $F$ to get a candidate module. However one can apply the special adjoint functor theorem. Module categories are complete, well powered, and have a cogenerator. So $F$ has a left adjoint $H$. You can apply EilenbergWatts to $H$ to deduce that $H \simeq  \otimes M$ for a bimodule $M$. Then $F \simeq \mathrm{Hom}(M,)$.
In your case you have a leftexact additive functor $\mathrm{mod}_R \to \mathrm{Mod}_S$, where $\mathrm{mod}_R$ denotes the compact objects (finitely presented modules). It extends uniquely to a functor $\mathrm{Mod}_R = \mathrm{Ind}(\mathrm{mod}_R) \to \mathrm{Mod}_S$ preserving filtered colimits. If we make the additional assumption that the extended functor preserves all products, then by the above argument it is of the form $\mathrm{Hom}(M,)$. Since it commutes with filtered colimits, $M$ is moreover compact.
Here is a (stupid in some sense) counterexample.
Suppose $A$ is not trivial and $\kappa$ is a cardinal greater than the cardinality of $\textrm{Hom}_A (X, M)$ for all f.g. $A$modules $X$ and $M$. Then the functor $M \mapsto M^\kappa$ is left exact and additive but not representable by any f.g. $A$module: if it were represented by $X$, then $X$ would have to be nontrivial, but then $X^\kappa$ would have cardinality $> \kappa$, and hence have cardinality greater than $\textrm{Hom}_A (X, X)$ – a contradiction. It is represented by the nonf.g. $A$module $A^{\oplus \kappa}$, though.
I'm afraid I don't have much useful to say about the case where you have a left exact and additive subfunctor of a representable functor, other than the easy observation that any representation of a subfunctor would have to be a quotient of the original representing object.
$\DeclareMathOperator{\Set}{Set} \DeclareMathOperator{\El}{El} \DeclareMathOperator*{\colim}{colim} \newcommand{\fC}{\mathcal C}$By abstract nonsense, for any category $\fC$, a functor $F:\fC\to \Set$ is a colimit of corepresented functors. Explicitly, let $\El(F)$ be the category of elements of $F$, whose objects are pairs $(c,x\in F(c))$ and morphisms $(c,x)\to (d,y)$ are morphisms $f:c\to d$ in $C$ such that $F(f)(x) = y$. There is an evident functor $p:\El(f)\to C$ which forgets the additional datum $x$, and we have $F\cong \colim_{\El(F)^{op}} y\circ p$, where $y:C^{op}\to \operatorname{Fun}(C,\Set)$ is the (opposite) Yoneda embedding.
In your situation, the functor $F$ preserves finite products (by additivity) and pullbacks (by left exactness), hence all finite limits. In this case, the index category $\El(F)$ is cofiltered, i.e. any functor $G:I\to \El(F), i\mapsto (c_i,x_i)$ extends to a cone $G_\triangleleft:I_\triangleleft\to\El(F)$: Let $c = \lim_I p\circ i$ be the limit of the underlying diagram and $x\in F(c)$ be a preimage of $\{x_i\}_{i\in I}$ under the isomorphism $F(c)\cong \lim_I F\circ G\subset \prod_{i\in I} F(c_i)$, so that sending the cone point to $(c,x)$ and the cone legs to the lifts of the projections from the limit $c\to c_i$ is the required extension. (Note that if $\mathcal C$ has all products and $F$ preserves them, you can take an arbitrary small category for $I$. If you could take $\El(F)$ itself, this means that $\El(F)$ has an initial object $(c_0,x_0)$, and unraveling the definitions shows that $F$ is corepresented by $c_0$. The adjoint functor theorem in Dan Petersen's answer essentially works this way.)
On the other hand, since colimits in functor categories are computed objectwise and filtered colimits commute with finite limits in Sets, any filtered colimit of corepresented functors preserves finite limits. So abstractly, the best thing you can hope for is a proobject representing your functor, i.e. a diagram $C:I\to \fC$ from a cofiltered category such that $F(c') \cong \colim_{I^{op}} \fC(C(i),c')$. Here you can always take $I$ to be the opposite of a cardinal, considered as a poset, by choosing a cofinal subset of $\El(F)$. For example, the functor $V\mapsto V^{\oplus\infty}$ is corepresented by the proobject with $I = \mathbb N^{op}$ and $C(n) = k^n$, with the maps $C(n+1)\to C^n$ given by projection. On the other hand, in Zhen Lin's example the category of elements $\El(F)$ has cofinality $\kappa$, which is bigger than the cofinality of any slice category, so that the functor can not be representable.
Restrict now to the setting in the question, i.e. $\mathcal C$ is the category of finitelygenerated modules over an Artin algebra $R$. We know that $F = \colim_{I^{op}} y(M_i)$ is a subfunctor of a corepresentable functor $y(M)$. This means that there are compatible maps $f_i:M\to M_i$. The submodules $\ker f_i$ of $M$ form a descending chain; since $R$ is Artinian and $M$ finitely generated, they stabilize at a finite level $i$. Replacing $M$ by $M/\ker f_i$ and $I$ by $I_{i/}$, we may assume that all the maps $f_i$ are injective. Similarly, the images of the maps $M_{i'}\to M_i$ form a descending chain of submodules of $M_i$, and therefore stabilize at a finite stage. Let $\{M'_i\}_{i\in I}$ be the proobject of these images, and $F' = \colim_{I^{op}} y(M'_i)$, so that we get a factorization $F\Rightarrow F'\Rightarrow y(M)$. The first map is an isomorphism (injectivity and surjectivity can be checked by lifting elements a finite amount through the colimit). Replacing $\{M_i\}_{i\in I}$ by $\{M'_i\}_{i\in I}$, we may assume that all transition maps are surjective. Thus all transition maps in the colimit $F'(M') = \colim_{I^{op}} \fC(M_i,M')$ are injective, so that $y(M_i)$ is a subfunctor of $F$ and hence of $y(M)$, which is equivalent to the maps $M\to M_i$ being surjective. Together with injectivity, this shows that we may assume that $\{M_i\}_{i\in I} = \{M\}_{i\in I}$ is a constant proobject, which is equivalent to corepresentability of $F$.