One way to make the subgroup $\mathrm{Sp}(n)\cdot\mathrm{Sp}(1)\subset\mathrm{SO}(4n)$ explicit is to think of $\mathbb{R}^{4n}$ as $\mathbb{H}^n$, i.e. as columns of quaternions of height $n$, where $\mathbb{H}\simeq\mathbb{R}^4$ is the ring of quaternions. Then $\mathrm{Sp}(n)$ can be thought of as the group of $n$-by-$n$ quaternion matrices $A$ that satisfy $A^\ast A = \mathrm{I}_n$, where $A^\ast$ is the conjugate transpose of $A$. This group then acts on $\mathbb{H}^n$ on the left by the usual matrix multiplication $A\cdot x = Ax$, for $A\in\mathrm{Sp}(n)$ and $x\in\mathbb{H}^n$. This action preserves the positive definite quadratic norm $\|x\|^2= x^\ast x$, so $\mathrm{Sp}(n)$ is exhibited as a subgroup of $\mathrm{SO}(4n)$.

This action is irreducible (it acts transitively on the unit sphere in $\mathbb{H}^n$), and its commuting ring is $\mathbb{H}$, thought of as scalar multiplication on the *right*:
$$
A\cdot (x q) = (A\cdot x) q
$$
for $q\in\mathbb{H}$ (which is true because quaternionic multiplication is associative). Remembering that $\mathrm{Sp}(1)$ is the group of unit quaternions, you can now see how $\mathrm{Sp}(n)\times\mathrm{Sp}(1)$ can act on $\mathbb{H}^n$, just use the rule
$$
(A,q)\cdot x = A\ x\ \bar q
$$
(the conjugation is important, because, otherwise, this won't be a left action). This action is not effective, because $(-A,-q)$ acts the same way as $(A,q)$, so you actually get a faithful representation of $\mathrm{Sp}(n)\cdot\mathrm{Sp}(1)\simeq \bigl(\mathrm{Sp}(n)\times\mathrm{Sp}(1)\bigr)/(\lbrace \pm\mathrm{I}_n,\pm 1\rbrace)$ into $\mathrm{SO}(4n)$.

The commuting ring of this representation is just $\mathbb{R}$, so there is no torus in $\mathrm{SO}(4n)$ that commutes with this action, which answers your second question.