Why is the Killing form of $\mathfrak{g}$ restricted to a subalgebra $\mathfrak{a} \subset \mathfrak{g}$ not the Killing form of $\mathfrak{a}$? I know that the Killing form of $\mathfrak{g}$ restricted to an ideal $I \subset \mathfrak{g}$ is just the Killing form of $I$.
However, what happens in general if we relax the conditions and just consider the Killing form restricted to a subalgebra $\mathfrak{a} \subset \mathfrak{g}$?

The question 'what happens' is  best answered by looking at an example, so the first question is: where do we find examples of subalgebras that are not ideals? 
Here is a class of examples. consider the real Lie algebra $\mathfrak{g} = \mathfrak{sl}_n$ of traceless $n$-by-$n$ matrices. It has a Cartan decomposition $\mathfrak{g} = \mathfrak{k} \oplus \mathfrak{p}$ where $\mathfrak{k}$ is the $(n(n-1)/2)$-dimensional subalgebra of anti-symmetric matrices and $\mathfrak{p}$ the $(n(n+1)/2 - 1)$-dimensional subspace (not algebra) of symmetric matrices. Inside $\mathfrak{p}$ there is a unique-up-to-conjugation maximal abelian subalgebra $\mathfrak{a}$ which we might as well (and, following convention, will) take to be the $n-1$-dimensional abelian subalgebra of diagonal matrices.
The nice thing about this set-up (well there actually many even nicer things about it, that is why these names are standard, but well) is that neither $\mathfrak{k}$ nor $\mathfrak{a}$ is an ideal in $\mathfrak{g}$.
Let's look at $\mathfrak{a}$ first. It is abelian so the killing form of $\mathfrak{a}$ itself is $0$. However, the restriction of the killing form of $\mathbb{g}$ to $\mathfrak{a}$ is not. We can decompose $\mathfrak{g}$ as the direct sum of $\mathfrak{a}$ and a bunch of root spaces for the adjoint action of $\mathfrak{a}$ (in other words: we can treat $\mathfrak{a}$ as a cartan subalgebra, this is because $\mathfrak{g}$ is split) and so we find that $ad(A)$ for some element $A \in \mathfrak{a}$ is diagonal with $n-1$ zeroes, $n(n-1)/2$ entries equal to $2$'s and $n(n-1)/2$ entries equal to $-2$.
It follows that $(A, A) = 4n(n-1)$ which is rather different from $0$. Here $(.,.)$ denotes the killig form of $\mathfrak{g}$. $(A_1, A_2)$ for different elements of $\mathfrak{a}$ yield multiples of 4 closer (in absolute value) to zero, but is zero only if $A_1 = -A_2$.
The case of the subalgebra $\mathfrak{k}$ is rather different, but I have to run now, maybe I come back to it later, or you can try it yourself. 
