First: The definition of Lie groups and a Lie algebras can vary depending on who you ask.
Given a (matrix) Lie group $G$ (so $G\subseteq GL_n(\mathbb{C})$), the Lie algebra of $G$ is the set
$$
\mathfrak{g} = \{X \in M_n(\mathbb{C}) \mid e^{tX} \in G \; \forall\; t\in \mathbb{R}\}.
$$
This is clearly then a real vector space. But it isn't necessarily a complex vector space. We always get a real Lie algebra. We only get a complex Lie algebra if $iX\in \mathfrak{g}$ for all $X\in \mathfrak{g}$. So just because the entries are complex, doesn't mean that the Lie algebra is complex.
Specifically considering $\mathfrak{sl}_2(\mathbb{C})$ you get the set of $2\times 2$ matrices with complex entries of trace zero. Again this is automatically a real vector space, but we can try to check if it is a complex vector space. We check that
$$
i\pmatrix{a & b \\ c & -a} = \pmatrix{ia & ib \\ ic & -ia}.
$$
This again has trace zero, so indeed $\mathfrak{sl}_2(\mathbb{C})$ is a complex Lie algebra. We usually then call the Lie group complex if the Lie algebra turns out to be a complex Lie algebra. Other complex Lie groups are $GL_n(\mathbb{C}), SL_n(\mathbb{C})$, $SO_n(\mathbb{C})$, and $Sp_n(\mathbb{C})$.
Another example: Consider the Lie group $SU(n)$ of all $n\times n$ unitary matrices (with entries from $\mathbb{C}$) with determinant $1$. In this case you can find that the Lie algebra $\mathfrak{su}(n)$ is the space of all $n\times n$ complex matrices $X$ where $X^* = -X$ ($*$ being complex conjugate transposed) and with trace $0$. This is not a complex Lie algebra, but only a real Lie algebra. So $SU(n)$ is not a complex Lie group.
Hopefully I didn't say anything wrong.
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
A matrix tangent to $SL(n)$ can be given by a smooth curve $$ c:]-\epsilon;+\epsilon[ \rightarrow M_n \quad \operatorname{det}(c(t))=1 \text{ for all t and } c(0)=E_n $$ det is multilinear and therefore has a derivative that is easy to compute $$ 0=\operatorname{trace}(c'(0)) $$