Why is $\mathbb{H}\otimes\mathbb{H}\cong\text{End}_\mathbb{R}\mathbb{H}$?

I guess by $\text{End}_{\mathbb{R}}$ you mean endomorphisms as an $\mathbb{R}$-vector space. In that case, $\mathbb{H}$ acts as endomorphisms of itself (as an $\mathbb{R}$-vector space) in two ways: left multiplication and right multiplication. That is, we have an $\mathbb{R}$-bilinear map $$\mathbb{H} \times \mathbb{H} \ni (q_1, q_2) \mapsto (x \mapsto q_1 x q_2) \in \text{End}_{\mathbb{R}}(\mathbb{H}).$$

The second action of $\mathbb{H}$ is a right action and not a left action, but we can fix this by conjugating to get $$\mathbb{H} \times \mathbb{H} \ni (q_1, q_2) \mapsto (x \mapsto q_1 x \overline{q_2}) \in \text{End}_{\mathbb{R}}(\mathbb{H}).$$

By the universal property of the tensor product (of algebras!) this gives an algebra map $$\mathbb{H} \otimes_{\mathbb{R}} \mathbb{H} \to \text{End}_{\mathbb{R}}(\mathbb{H})$$

and the question is to show that this map is an isomorphism. Since both algebras have dimension $16$, it suffices to show that this map is either surjective or injective. Surjectivity follows from the Jacobson density theorem, for example.