How to find the radical of an acyclic quiver and without relations? what is the recipe?
For example suppose $Q$ is a quiver with two vertices and two arrows from vertex $2$ to vertex $1$. Now suppose $M$ is a representation given by $K^3 \leftleftarrows K^4$ where K^3 corresponds to the vertex $1$ and $K^4$ to the vertex $2$ (here $K$ of course is a field). Here $\leftleftarrows$ denotes two distinct maps, the two projections $f\colon K^4 \to K^3\colon f(a,b,c,d)=(a,c,d)$ and $g\colon K^4 \to K^3\colon g(a,b,c,d)=(b,c,d)$.
How do you find the radical corresponding to the representation $M$?
Added After reading Assem's book I'm still confused. Do you need to take the image of the corresponding maps?
So for example there are no upcoming arrows to vertex $2$ so the corresponding vector space attached to vertex $2$ is the zero space. Now the are two arrows from vertex $2$ to $1$. We look at the images of the maps. Both image (being surjective) are $K^3$.
Now we take the sum of submodules: $K^3 + K^3 = K^3$.
Is this how you proceed?
You want to find the radical of a representation $M$, in [ASS]-notation this is $\operatorname{rad} M$.
By Proposition 3.7 (d) in this book, you have $\operatorname{rad}M=M\operatorname{rad}A$. For an acyclic quiver, its Jacobson radical $\operatorname{rad}A$ coincides with the ideal generated by the arrows, called $R_Q$ in [ASS].
Now this ideal is spanned by the arrows, and you are given a basis of the representation. To get a generating set for the radical of the module, you now only need to compute each product of an arrow by a basis element of your representation. By the equivalence of representations and modules, this is just applying the map on the arrow to the basis vector.
Your technique is of course also correct, you take the images of all the maps on the arrows.
As the result, in your case you get $0\to K^3$ as the radical of $K^4\to K^3$. Note that $\operatorname{rad}(M)$ is still a representation of $Q$.