Your notation is unconventional, but I'll humor you,
$$|3/2,3/2\rangle = |1, 1\rangle \otimes |1/2, 1/2\rangle.
$$
In simpler, intuitive notation, (vectors of the right space into vector components of the left one), this is just
$$
\begin{pmatrix}1\\0\\0\\0\end{pmatrix}\oplus\begin{pmatrix}0\\0\end{pmatrix}= \begin{pmatrix}1\\0\\0\end{pmatrix}\otimes \begin{pmatrix}1\\0\end{pmatrix}= \begin{pmatrix}1\\0\\0\\0\\0\\0\end{pmatrix} ,
$$
reminding you of the "accidental" Clebsches being 1 and 0 respectively for the reduced spin subspaces. Had you chosen m=1/2 on the left, life would have been dramatically different, since the vector spaces on either side are really 6-dimensional, 3/2 ⊕ 1/2 = 1 ⊗ 1/2 . Normally, you go from the r.h.side to the l.h.side which is special in your case.
But since you are asking for your extreme, unrepresentative, example, for linear angular-momentum operators, such as $\hat j_0, \hat j_{\pm}$, you'd have, indeed, in this case (only),
$$
\Delta(\hat j_i)= (\mathbb{1} \otimes \hat{j_i} + \hat{j_i} \otimes \mathbb{1}),
$$
so
$$\hat{j_i}|3/2, 3/2\rangle = (\mathbb{1} \otimes \hat{j_i} + \hat{j_i} \otimes \mathbb{1})(|1, 1\rangle \otimes |1/2, 1/2\rangle ).
$$
In a less sadistic , conventional, notation, e.g. for $\hat j_0$,
$$
(3/2)\begin{pmatrix}1\\0\\0\\0\end{pmatrix}\oplus\begin{pmatrix}0\\0\end{pmatrix}= \begin{pmatrix}1\\0\\0\end{pmatrix}\otimes (1/2)\begin{pmatrix}1\\0\end{pmatrix}+ \begin{pmatrix}1\\0\\0\end{pmatrix}\otimes \begin{pmatrix}1\\0\end{pmatrix} =
(3/2) \begin{pmatrix}1\\0\\0\end{pmatrix}\otimes \begin{pmatrix}1\\0\end{pmatrix}.
$$
Although correct, this is somewhat misleading, since it does not dramatize the need for Clebsching: basis changes on the left to disentangle the invariant subspaces.
And likewise for the other two su(2) generators. Of course, for $\hat j_+$ on the l.h.side and $\Delta(\hat j_+)$ on the r.h.side you get 0. For $\Delta(\hat j_-)$,
$$
\sqrt{3/2}\begin{pmatrix}0\\1\\0\\0\end{pmatrix}\oplus\begin{pmatrix}0\\0\end{pmatrix}= \begin{pmatrix}1\\0\\0\end{pmatrix}\otimes (1/\sqrt{2})\begin{pmatrix}0\\1\end{pmatrix}+ \begin{pmatrix}0\\1\\0\end{pmatrix}\otimes \begin{pmatrix}1\\0\end{pmatrix} .
$$
On the r.h.side you got $(0, 1/\sqrt{2}, 1,0,0, 0)^T$, which reduced to the l.h.side in a different (Clebsched) basis.
However, as emphasized in a linked answer, if the operator $\hat O$ is a Casimir invariant, instead, the quadratic in generators labelling the representation (spin), then the above formula fails, dramatically. You must square the Δs of each generator, and sum them together, giving you a more elaborate expression, since there are cross terms. Your notation would not serve you then. To preserve your sanity, I'd write the corresponding conventional expression,
$$
(\mathbb{1} \otimes \hat{j_i} + \hat{j_i} \otimes \mathbb{1})(\mathbb{1} \otimes \hat{j_i} + \hat{j_i} \otimes \mathbb{1})\left (\begin{pmatrix}1\\0\\0\end{pmatrix}\otimes \begin{pmatrix}1\\0\end{pmatrix}\right )
$$
on the right, which would Clebsch, on the left, to 15/4 acting on any quartet (any spin 3/2 state), and 3/4 on the doublet (but not the trivial null spin 1/2 state here).
In undergraduate addition-of-angular-momentum language:
$$
|3/2 ~ 3/2\rangle = |1 1;1/2 1/2\rangle \\
\hat {j}_0: \qquad 3/2~ |3/2 ~ 3/2\rangle =(1+1/2) |1 ~1;1/2~ 1/2\rangle \\
\hat {j}_+:\qquad \qquad\qquad\qquad 0=0 \qquad\qquad\qquad\\
\hat{ j}_-: ~~ \sqrt{3/2}~ |3/2 ~ 1/2\rangle = |1 ~0;1/2 ~1/2\rangle +(1/\sqrt{2}) |1 ~1;1/2 ~-1/2\rangle \\
\hat {j}^2: \qquad 15/4 ~ |3/2 ~ 3/2\rangle =( 15/4) |1 ~1;1/2 ~1/2\rangle
. $$
