I have a finite dimensional $k$-vector space $V$ and a $k$-algebra $R$.
I want to prove the following sequence of isomorphisms.
$$\text{End}_R(V\otimes R)\cong \text{Hom}_k(V,V\otimes R)\cong \text{Hom}_k(V\otimes V^*, R)\cong \text{Hom}_k(\text{End}(V),R)\cong \text{Hom}_{\text{Alg}}({\cal{O}}(\text{End}(V)),R)$$
Where ${\cal{O}}(\text{End}(V))$ is actually the coordinate algebra and so it is $k[X_{11},\cdots,X_{nn}]$
For the first isomorphism, take $\phi(\alpha)(e_i) = \alpha(e_i\otimes1)$
For the rest I have no idea
First you should specify what kind of algebras you are talking about. Like unital, commutative, ...
Now let me note that there is an isomorphism $$V^* \otimes W \cong \text{Hom}_k(V,W)$$ thus $$\text{Hom}_k(V,V\otimes R)\cong \text{Hom}_k(V,\text{Hom}_k(V^*,R))$$ and by universal property of the tensor product we have $$\text{Hom}_k(V,\text{Hom}_k(V^*,R))\cong \text{Hom}_k(V\otimes V^*, R).$$
This gives the second isomorphism.
For the thrid just note that $$V\otimes V^* \cong V^*\otimes V \cong \text{End}_k(V).$$
So we are left with the last one. I am not sure here. So $\mathcal{O}(\text{End}(V))$ is just the free algerba on a basis of $\text{End}_k(V)$ if I get it right? Since $V$ has dimension $n$(?) we can choose a base $(X_{ij})_{i,j \le n}$ of $\text{End}_k(V)$ and then $\mathcal{O}(\text{End}(V))$ is given by $k[X_{ij}]_{i,j \le n}$.
But a morphism from a polynomial $k$-algebra $k[X_{ij}]_{i,j \le n}$ to an $k$-algebra $R$ given by $k$-linear maps from $k[X_{ij}]$ to $R$ for each $i,j$. Now since $\text{Hom}_k(\text{End}(V),R)$ is fixed on a basis $(X_{ij})_{i,j \le n}$ too you can find a bijection.
There are now serval problems with it. Is $\text{Hom}_{\text{Alg}}({\cal{O}}(\text{End}(V)),R)$ a vector space? How does the last isomorphism look like? The first three were all canonical. I.e. you didn't made any specific choices and they are all $k$-linear. The last one is only a bijection and it depends on the choice of the basis.