I would like to understand the equivalence of the 2 facts that hom is left exact in the sense that it carries short exact sequences to left exact sequences and that it preserves all small existing limits. I know the proof of the former and now I would like to carry it over to the second fact thereby establishing that in fact it preserves all small limits.
EDIT perhaps to make sense of this equivalence is to confine the hom to additive (or abelian ?) categories and additive functors. If this is still too general, take $\operatorname{hom}(M,-)=\operatorname{Mod}_R(M,-).$
$\newcommand{\A}{\mathscr{A}}\newcommand{\B}{\mathscr{B}}\newcommand{\Ke}{\mathsf{Ke}}$This question has been answered before on MSE for specific cases, e.g. the category of modules. I didn't find a general answer though, so I'll put one here.
I assume the following context: $\A,\B$ are Abelian categories, $T:\A\to\B$ is some additive functor.
We want to show: $T$ is left-exact iff. it is finitely continuous. The additivity requirement is necessary since it implies $T$ preserves all finite biproducts. It is well-known that all you need to build limits are all products and all equalisers; the statement about being left-exact pertains to preserving equalisers, and the additivity to preserving finite (bi)products. These are two separate claims.
Note: I don't actually know what you mean by a Hom-functor. Furthermore, the preservation of kernels only implies preservation of finite equalisers, not of infinite ones (I don’t think, anyway) so you’d need more hypotheses to have $T$ wholly continuous.
The goal is to show that, for any $f\in\A(a,b)$, and kernel arrow $\ker f:\Ke(f)\to a$, $T(\ker f):T(\Ke(f))\to T(a)$ is a kernel arrow for $T(f)$. A kernel is just an exact sequence: $$\Ke(f)\overset{\ker f}{\longrightarrow}a\overset{f}{\longrightarrow}b$$
I'll borrow the image from this answer:
Allow $A_1=\Ke(f),\,A_2=a,\,A_3=b$. $C_{1,2}\to A_{1,2}$ is the kernel of $A_{1,2}\to A_{2,3}$, and $C_3\to A_3$ completes the mono-epi factorisation, and $A_3\to C_3$ is $\operatorname{coker} f$. I invite you to check for yourself that the diagonals involving the $C_\bullet$ are / can be naturally filled out to be short exact sequences. So by hypothesis, $T$ lifts this diagram to a similar diagram in $\B$ where all the diagonals are left exact sequences.
Let's label $q_1=T(C_1)\to T(A_1)$, $q_2=T(A_1)\to C_2$, $q_3=T(C_2)\to A_2$, $q_4=T(A_2)\to T(C_3)$, $q_5=T(C_3)\to T(A_3)$, $q_6=T(A_3)\to T(C_3)$. The assumption on $T$ is that $q_1=\ker q_2$, $q_3=\ker q_4$, and $q_5=\ker q_6$. $T(f)=q_5q_4$: $q_5q_4\circ\alpha=0\iff q_4\circ\alpha=0$ as $q_5$ is monic, so $\ker(T(f))\equiv\ker q_4$. But that's just $q_3$!
We know that $T(\ker f)=q_3q_2$. However, $q_1=T(\ker(\ker f))$ by definition of the arrow $C_1\to A_1$: $\ker\ker f\equiv0$, so in fact the arrow $A_1\to C_2$ is an isomorphism. Then $q_2=T(A_1\to C_2)$ is an isomorphism, so $T(\ker(f))=q_3q_2$ means that, up to equivalence: $$\ker(T(f))=q_3\equiv q_3q_2=T(\ker(f))$$
Thus $T$ is an additive, kernel-preserving functor. The equaliser of a parallel pair $f,g:a\to b$ is the kernel of $f-g:a\to b$, so $T$ carries this kernel to a kernel of $T(f-f)=T(f) -T(g)$ in $\B$ as $T$ is additive and preserves kernels: thus, we have an equaliser of $T(f),T(g)$, which was preserved. That is, $T$ is continuous for all finite products and equalisers, thus continuous for all finite limits. $\blacksquare$.