Let $Q$ be an acyclic quiver, and $V = (V_i, V_a)$ a representation of $Q$. I want to show that $$\text{soc}(V) = \bigoplus_{i \in Q_0} \left(\bigcap_{a \in Q_1, s(a) = i} \ker(V_a) \right), \hspace{3mm} \text{rad}(V) = \bigoplus_{i \in Q_0} \left(\sum_{a \in Q_1, t(a) = i} \text{im} (V_a) \right).$$ I think I succeeded in showing the equality for soc:
Let $S \subseteq V$ be a simple submodule. By some previous result we know that for an acyclic quiver, the simple modules are (up to isomorphism) of the form for $S_j = K$ for one $j \in Q_0$, $S_i = 0$ for $i \neq j$, and all $S_a = 0$. If $j$ is not the starting point of any arrow, then we defined $\bigcap_{a \in Q_1, s(a) = j} \ker(V_a) := V$, so in this case we obviously have $S \subseteq \bigcap_{a \in Q_1, s(a) = j} \ker(V_a)$. On the other hand, if $a \in Q_1$ with $s(a) = j$, then $V_a(S_j) \subseteq S_{t(a)} = 0$ since $S$ is a submodule, hence $S_j \in \ker(V_a)$ for any such $a$, so especially $S \subseteq \bigcap_{a \in Q_1, s(a) = j} \ker(V_a)$. Since $\text{soc}(V)$ is spanned by such $S$, we get "$\subseteq$". For the other direction, we note that for $j \in Q_0$, any subspace of $\bigcap_{a \in Q_1, s(a) = j} \ker(V_a)$ is a submodule, so either the intersection is $0$ or semisimple, and $\text{soc}(V)$ contains all semisimple modules, hence "$ \supseteq$". Is this correct so far?
Now I wanted to show the equality for $\text{rad}(V)$, however I am having a few problems. First off, I don't really know how maximal submodules of $V$ look like. If we would consider the quiver $1 \rightarrow 2$, and let $V$ be the representation $K^3 \rightarrow K^3$, where the map is just the identity, then what are the maximal submodules? Intuitively I would say, that $K^2 \rightarrow K^3$ is a maximal submodule, and I would further say, that it is the only maximal one, so $\text{rad}(V) = (K^2 , K^3)$. The submodules are all of the form $K^n \rightarrow K^m$, where $0 \leq n,m \leq 3$ and $n \leq m$, and it looks like that all of them (the proper ones) are contained in $K^2 \rightarrow K^3$. However, when using the equation above, we should actually get $\text{rad}(V) = 0 \rightarrow K^3$, but I fail to see how.
I understand that for a "normal" module $V$, we say that a submodule $U \subset V$ is maximal, if $U \subseteq U'$ for a proper submodule $U' \subset V$ implies $U = U'$. Since every quiver representation is equivalent to a $KQ$-module, we hence consider $\bar{V} := K^3 \oplus K^3$ and look for submodules $\bar{U} \subset \bar{V}$ that are stable under the action $(x,y) \mapsto (0,x)$. Is that correct? But this again looks like as if $(K^2, K^3)$ is the maximal submodule. I don't know what I am doing wrong, could someone give me some advice?
Also, is there an easy example of a cyclic quiver $Q$ where the above equations do not hold? I am relatively new in representation theory and find it hard to think of examples, this would help a lot!
Maximal submodules are the kernels of nonzero maps to simple modules, and the radical is basically dual to the socle.
Consider your example of the identity map $K^3\to K^3$. There is no nonzero map to the simple $S_2: 0\to K$, but every linear map $f\colon K^3\to K$ gives a map to the simple $S_1: K\to 0$, and the kernel of this map is $\mathrm{Ker}(f)\to K^3$. So the maximal submodules are $U\to K^3$ for any two-dimensional subspace $U\leq K^3$. The radical is the intersection of all such, so is $0\to K^3$.
Thinking of this in terms of the map $(x,y)\mapsto(0,x)$, we again see that every pair of subspaces $M\leq N\leq K^3$ gives rise to a submodule $M\to N$ of $K^3\to K^3$. When is it maximal? We can always increase $N$ to $K^3$, and then increase $M$ to any two-dimensional subspace $U$ of $K^3$, so again the maximal submodules are the $U\to K^3$ with $\dim U=2$.
In general, when do we have a map $f\colon V\to S_i$? This is given by a map $f_i\colon V_i\to K$ satisfying the necessary conditions for arrows. These conditions are immediate for arrows starting at $i$, and so we only need the condition for arrows $a$ ending at $i$, in which case it becomes the condition that the image of $V_a$ is contained in the kernel of $f_i$, or alternatively the kernel of $f_i$ contains the images of all the $V_a$.
For cyclic quivers, there are many more simple modules. For example, for the Jordan quiver (one vertex, one loop), every one-dimensional module is simple, so we have simples indexed by the elements of $K$. (The path algebra here is just $K[t]$, and so we are just considering finite dimensional $K[t]$-modules, or vector spaces equipped with an endomorphism).