The direct sum of $n$-copies of a $K$-vector space $V$ is a free $\operatorname{End}_{R}(V)$-module

If $n = 1$ (note that $n$ also has to be the dimension of $V$) then actually the other question doesn't apply, because it requires $n > 1$.

At any rate, let's check this. First note that by the definition of the direct sum of $R$-modules, the module action of $\operatorname{End}_K(V)$ on $V^n$ is just to act on each factor separately. That is, for any $\phi \in \operatorname{End}_K(V)$ and $v = (v_1, \ldots, v_n) \in V^n$ we have $$ \phi \cdot v := (\phi(v_1), \ldots, \phi(v_n)). $$ Ok, now given this definition we are supposed to concoct a basis for $V^n$ as a module over $\operatorname{End}_K(V)$. As a guess, let's pick a fixed basis $\mathcal{B} = \{b_1, \ldots, b_n\}$ of $V$ as a module over $V$. We'll use this for our basis for $V^n$: let's try the set $\{(b_1, \ldots, b_n)\} \subset V^n$ containing just $1$ element.

There are two things to check: first, $\operatorname{End}_K(V)$-linear independence. Thus suppose that $$ \phi \cdot (b_1, \ldots, b_n) = 0 $$ for some $\phi \in \operatorname{End}_K(V)$ (note that the definition of linear independence for a set containing a single element is particularly simple). We need to check that $\phi = 0$. To see this now just note that the above equation implies that $\phi(b_i) = 0$ for all $i$. But then $\phi$ is a linear map $V \to V$ which maps a basis of $V$ to zero! Hence $\phi = 0$, as desired.

It just remains to check that the set $\{(b_1, \ldots, b_n)\} \subset V^n$ is $\operatorname{End}_K(V)$-spanning. Thus now let $v = (v_1, \ldots, v_n) \in V^n$ be arbitrary. We need to produce $\phi \in \operatorname{End}_K(V)$ such that $\phi \cdot (b_1, \ldots, b_n) = (v_1, \ldots, v_n)$. But again since the set $\{b_1, \ldots, b_n\}$ is a basis for $V$ we may simply define a linear map $\phi : V \to V$ by declaring that $\phi(b_i) = v_i$ for all $i$. Then indeed $\phi \cdot (b_1, \ldots, b_n) = (v_1, \ldots, v_n)$, as desired. This completes the proof.

In summary, we have shown that $V^n$ is a free $\operatorname{End}_K(V)$-module with a basis consisting of one element. This actually (always) means that $V^n$ and $\operatorname{End}_K(V)$ are isomorphic $\operatorname{End}_K(V)$-modules. (Note that any ring $R$, including $R = \operatorname{End}_K(V)$, can be considered as an $R$-module over itself.) The interpretation is that (after choosing a basis of $V$) elements of $V^n$ can be thought of as $n \times n$ matrices, where each of the $n$ summands in $V^n$ is to be thought of as a matrix column. From this perspective, the module action of $\operatorname{End}_K(V)$ on $V^n$ is then just matrix multiplication.