Well for a start we cannot write every element of $G$ as a matrix nor as an exponential. Lie algebras can always be written as matrices by Ado's theorem but that is not true for Lie groups. Of course we should really strive for a proof without matrices anyway since writing something in terms of matrices requires us to not only pick a representation but pick a basis of that representation. We would really need to show that our proof did not depend on these choices and even then a more direct proof is cleaner. Then the second point. Even if we restrict ourselves to only considering connected Lie groups the exponential map (which depends on which Lie group of our Lie algebra we are considering and is not in general just the matrix exponential) will, in general, fail to be surjective unless, for example, the group is compact or nilpotent. Instead the image of the exponential generates the group (or more generally the connected component of the identity).
The results you are referring to are simply showing the derivative of a Lie group homomorphism is a Lie algebra homomorphism $\mathfrak{g}\to \mathfrak{h}$ (number 2) and that this is equivariant with respect to the natural adjoint actions on $\mathfrak{g}$ and $\mathfrak{h}$ (number 1).
So to answer your first question: we are not really, except for the adjoint representation. Using matrices would be invoking representation theory. For your second, I'm not sure. These are somewhat basic facts of the Lie algebra-Lie Group correspondence. I don't think they rely on any overly technical parts of Lie theory as long as you are comfortable dealing with Lie Groups/Algebras in the abstract.