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I proved that an additive functor $F: R-Mod\to S-Mod$ is left exact if and only if it preserves kernels.

Is there an abstract nonsense way of deriving the dual statement that $F$ is right exact if and only if it preserves cokernels from the statement that I already proved?

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    You need to first replace the categories modules by abelian categories and prove a similar statement. Then, since the opposite of an abelian category is abelian you obtain the desired dual statement by applying the more general fact to $F^{op}$. – 2017-01-04

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I would say it depends on your proof and what you are willing to put into the dualizing argument.

What you need is that if $$ 0 \to A \xrightarrow{k} B \xrightarrow{f} C $$ is exact, if and only if $k$ is a kernel.

Now the first problem that occurs: what is exactness at $A$ and $B$? In the category of modules this is clear, but if you use the opposed category you need to make up a new definition. So you can make up a categorical definition of the image and then use this for the proof. You are only allowed to use categorical terms. Then you can simply dualize the statement.

The other way would be to say the categories can be restricted to small abelian ones and use Mitchell's embedding theorem. Then you can use your proof and it does not depend on how the proof looks like and dualize.

I would prefer the first approach because you could see there in which generality this result holds. Does it hold in additive categories? On the other hand working with elements is nice.