I'm looking for an intuitive explanation of the meaning of the Lie algebra for a matrix Lie group from a differential geometry perspective.
Right now, the procedure I've been following is using the Regular Level Set Theorem and showing that for some surjective function $F : M \rightarrow N$, if $c$ is a regular value of $F$ then $F^{-1}(c) := \{ p \in M : F(p) = c\}$ is an embedded submanifold and $ Lie(F^{-1}(c)) = T_ F^{-1}(c) \cong Ker(F_{*, })$.
I've been able to show that the lie algebra of the orthogonal group:
$$ O(n) = \{ A \in M_{n \times n}(\mathbb{R}) : AA^T = \} $$ is $Skew(n, \mathbb{R}) = \{ A \in M_{n \times n}(\mathbb{R}) : A^T = - A\}$, and that the Lie algebra of the Special Linear group:
$$ SL(n, \mathbb{R}) = \{ A \in M_{n\times n} : det(A) = 1 \} $$
is the set of trace $0$ matricies.
The Lie algebras I'm getting seem completely arbitrary to me and I can't really see what the relation is with the original Lie group!
Maybe you don't know there's an exponential map that takes the Lie algebra to the Lie group: $$\exp \colon \mathfrak{g} \to G$$
In general, this gives a diffeomorphism from a neighborhood of 0 in the Lie algebra to a neighborhood of the identity in $G$.
For your concrete matrix groups, we can use this to compute the Lie algebras easily. Let's look at the case of $O(n)$. Say $X$ is in the Lie algebra. We have the 1-parameter subgroup $e^{tX}$ which satisfies $e^{tX}e^{tX^T} = I$. Differentiating at $t = 0$, we get $X+X^T = 0$ as you discovered. You can do the $SL(n)$ case just as easily.