As you suspect, neither $\mathfrak{su}(4)$ nor $\mathfrak{so}(6)$ is isomoprhic to $\mathbb R^{15}$ as a Lie algebra, exactly because the first two have nontrivial brackets but the bracket on the last is zero.
Note that $\mathfrak{su}(4)$ is defined in terms of its action on $\mathbb C^4$, and $\mathfrak{so}(6)$ is defined in terms of its action on $\mathbb R^6$. So the best way to show that $\mathfrak{su}(4)$ are $\mathfrak{so}(6)$ is to make the first act on $\mathbb R^6$ or the second on $\mathbb C^4$, and then to check that these actions are the ones desired.
I find the former easier, since it extends to an action of $\mathrm{SU}(4)$ on $\mathbb R^6$, whereas $\mathrm{SO}(6)$ does not act in the desired way on $\mathbb C^4$. The trick is to notice that $\binom42 = 6$. We take the $\mathrm{SU}(4)$ action on $\mathbb C^4$, and use it to act on $\mathbb C^4 \wedge_{\mathbb C} \mathbb C^4 \cong \mathbb C^6$ by $g(v\wedge w) = gv \wedge gw$ for $g\in \mathrm{SU}(4)$ and $v,w \in \mathbb C^4$; the infinitesimal version of this is $x(v\wedge w) = xv \wedge w + v \wedge xw$ for $x\in \mathfrak{su}(4)$.
Of course, the action of $\mathrm{SU}(4)$ on $\mathbb C^4$ extends to an action of $\mathrm{SL}(4,\mathbb C)$. So I will temporarily work with it.
Define a pairing $\langle,\rangle$ on $\mathbb C^4 \wedge_{\mathbb C} \mathbb C^4 \cong \mathbb C^6$ by $\langle v_1\wedge w_1, v_2\wedge w_2 \rangle = \det(v_1,w_1,v_2,w_2)$, where $(v_1,w_1,v_2,w_2)$ denotes the matrix with rows $v_1,w_1,v_2,w_2$ — that this is well-defined follows from standard facts about the determinant. By definition, $\mathrm{SL}(4,\mathbb C)$ consists of all $\mathbb C$-linear automorphisms of $\mathbb C^4$ that preserve the determinant, and therefore the $\mathrm{SL}(4,\mathbb C)$ action on $\mathbb C^6$ preserves this pairing.
But the pairing is nondegenerate, also by standard facts about the determinant. Over $\mathbb C$, a vector space has a unique-up-to-isomorphism nondegenerate pairing. It follows that the action of $\mathrm{SL}(4,\mathbb C)$ on $\mathbb C^6$ factors through the action of $\mathrm{SO}(\mathbb C,6)$, where $\mathrm{SO}(\mathbb C,6)$ is the group of complex matrices preserving the pairing $\langle,\rangle$ (isomorphic to any other copy of such a group).
So, we have constructed a homomorphism $\mathrm{SU}(4) \to \mathrm{SL}(4,\mathbb C) \to \mathrm{SO}(6,\mathbb C)$. But the domain $\mathrm{SU}(4)$ is a compact group, and so its image must be compact (since the homomorphism is a continuous map of manifolds). It is a fact (but I don't remember how easy it is to prove) that every compact subgroup of $\mathrm{SO}(6,\mathbb C)$ is contained within a conjugate of $\mathrm{SO}(6,\mathbb R)$. We therefore get a map $\mathrm{SU}(4) \to \mathrm{SO}(6,\mathbb R)$.
Finally, it is not difficult to check that the action of $\mathfrak{sl}(4)$ on $\mathbb C^6$ constructed above has trivial kernel. Indeed, suppose that $x \in \mathfrak{sl}(4)$ acts trivially. Choose the standard basis $e_1,\dots,e_4$ of $\mathbb C^4$; it induces a basis $e_{12},e_{13},\dots,e_{34}$ on $\mathbb C^6$, where $e_{ii'} = e_i \wedge e_{i'}$ for $i<i'$. If the $(i,j)$th matrix entry for $x$ was $x_i^j$, so that $x(e_i) = \sum_j x_i^j e_j$ then $x(e_{ii'}) = x(e_i)\wedge e_{i'} + e_i \wedge x(e_{i'}) = \sum_j x_i^j e_j \wedge e_{i'} + \sum_{j'} x_{i'}^{j'} e_i \wedge e_{j'}$. For $x$ to act by zero, this sum would have to be zero for all values of $i,i'$. But since we are in four dimensions, for any $i,i'$, there is a $j \neq i,i'$, whence $e_j \wedge e_{i'}$ is independent of $e_i \wedge e_{j'}$ for any $j'$. Thus the only way for $x$ to act as $0$ on $\mathbb C^6$ is if $x_i^j = 0$ for all $i,j$.
Therefore the map $\mathfrak{su}(4) \to \mathfrak{so}(6)$ constructed above has trivial kernel. Since it is between two Lie algebras of the same (finite) dimension, it therefore must be an isomorphism.
Best Answer
The two Lie algebras are isomorphic over $\mathbb{C}$, but not over $\mathbb{R}$. The multiplications in $\mathfrak{sl}(2,\mathbb{R})$ are $[H,X]=2X,[H,Y]=-2Y$ and $[X,Y]=H.$ Here we have $\operatorname{ad}H=\operatorname{diag}(0,2,-2).$ Now the multiplications in $(\mathbb{R}^3,\times)$ are given by $[U,V]=W,[V,W]=U,[W,U]=V.$ If there was an isomorphism, then there would exist a vector $T:=uU+vV+wW$ which maps onto $H.$ Then $\operatorname{ad}T$ maps on $\operatorname{ad}H.$ You can calculate the eigenvalues of $\operatorname{ad}T$ which should include a complex one.