"I sometimes use the fact that if there are $a_{k}$ ways to choose
$k$ out of $n$ conditions and fulfill them..."
In this answer I choose for a setup starting with a set $X$ containing what
you call ways as elements. Each of the $n$ conditions corresponds with a subset
of $X$ that contains exactly the ways to fulfill the condition.
So let me introduce an index set $I$ with cardinality $n$ and the
collection $\left\{ A_{i}\mid i\in I\right\} $ where $A_{i}\subseteq X$
contains the ways that fulfill condition $i$.
For $J\subseteq I$ we define: $$A_{J}:=\bigcap_{i\in J}A_{i}$$ under
the convention that $A_{\varnothing}=X$.
Then $a_{k}$ mentioned above can be recognized as: $$a_k=\sum_{J\subseteq I\wedge\left|J\right|=k}\left|A_{J}\right|$$
In the special case where the cardinality of $J$ is determining for the
cardinality of $A_{J}$ and $b_{k}:=\left|A_{J}\right|$ whenever $\left|J\right|=k$
we have the equality: $$a_{k}=\binom{n}{k}b_{k}$$ which is also mentioned
in your question.
Finally for nonnegative integer $j$ we define: $$U_{j}:=\left\{ x\in X\mid\sum_{i\in J}\mathbf{1}_{A_{i}}\left(x\right)=j\right\}\text{ and }u_j:=|U_j| $$
So that we will have $x\in U_{j}$ iff $x$ is a way that fulfills exactly
$k$ of the conditions, and - as you said in your question - we will have: $$u_{j}=\sum_{k=j}^{n}\left(-1\right)^{k-j}\binom{k}{j}a_{k}\tag1$$
Uptil now I only sketched the setup and now it is time to find combinatorial
proof of this.
Lemma: If $S$ is a finite and non-empty set then: $$|\{T\in\mathcal P(S)\mid |T|\text{ is odd}\}|=|\{T\in\mathcal P(S)\mid |T|\text{ is even}\}|$$
Proof: Straightforward.
Theorem: For every nonnegative integer $j$ we have: $$\mathbf{1}_{U_{j}}+\sum_{J\subseteq I\wedge\left|J\right|-j\text{ is odd}}\binom{\left|J\right|}{j}\mathbf{1}_{A_{J}}=\sum_{J\subseteq I\wedge\left|J\right|-j\text{ is even}}\binom{\left|J\right|}{j}\mathbf{1}_{A_{J}}$$
Before proving the theorem let us first have a look at its impact.
If some measure $\mu$ is involved and we are dealing with measurable sets
then integration on both sides yields: $$\mu\left(U_{j}\right)+\sum_{J\subseteq I\wedge\left|J\right|-j\text{ is odd}}\binom{\left|J\right|}{j}\mu\left(A_{J}\right)=\sum_{J\subseteq I\wedge\left|J\right|-j\text{ is even}}\binom{\left|J\right|}{j}\mu\left(A_{J}\right)$$
If we take the counting measure then automatically the sets are measurable
and at first hand we find: $$u_{j}+\sum_{k=j\wedge k-j\text{ is odd}}^{n}\binom{k}{j}a_{k}=\sum_{k=j\wedge k-j\text{ is even}}^{n}\binom{k}{j}a_{k}$$
where $a_{k}$ and $u_{j}$ are defined as above.
If moreover the $a_{k}$ are finite then we can subtract without problems so that we arrive at: $$u_{j}=\sum_{k=j}^{n}\left(-1\right)^{k-j}\binom{k}{j}a_{k}\tag1$$
So $(1)$ follows immediately from the theorem and is not more
than a special case.
Surprisingly the theorem is not difficult to prove and
the only essential thing needed for that is the lemma.
Proof:
Let $x\in X$.
Then it is enough to prove that substituting $x$ on both
sides gives equal outcomes. For this let: $$I_{x}:=\left\{ i\in I\mid x\in A_{i}\right\} $$
and discern the following cases:
$\left|I_{x}\right|<j$ then we find $0+0=0$ on LHS and $0$ on RHS.
$\left|I_{x}\right|=j$ then we find $1+0=1$ on LHS and $1$ on RHS.
$\left|I_{x}\right|>j$ then we find: $$0+\sum_{J\subseteq I_{x}\wedge\left|J\right|-j\text{ is odd}}\binom{\left|J\right|}{j}\binom{\left|I_{x}\right|}{\left|J\right|}=\binom{\left|I_{x}\right|}{j}\sum_{J\subseteq I_{x}\wedge\left|J\right|-j\text{ is odd}}\binom{\left|I_{x}\right|-j}{\left|J\right|-j}$$
on LHS and: $$\binom{\left|I_{x}\right|}{j}\sum_{J\subseteq I_{x}\wedge\left|J\right|-j\text{ is even}}\binom{\left|I_{x}\right|-j}{\left|J\right|-j}$$
on RHS.
The factor $\binom{\left|I_{x}\right|}{j}$ on both sides can be striped
away and what remains is exactly the true statement that
the number of subsets of a set having the finite cardinality $\left|I_{x}\right|-j>0$
with odd cardinality equals the number of subsets with even cardinality. So the content of the lemma.
This completes the proof.
This is not very difficult combinatorial proof of what you call the general version of inclusion/exclusion and is built on not more than one very simple lemma.