Why the whole exterior algebra?

You might prefer to think of the exterior algebra as a graded algebra. A graded algebra is a monoid in the monoidal category of graded vector spaces, just as an algebra is a monoid in the monoidal category of vector spaces.

Wikipedia defines these objects as algebras (respectively, vector spaces) with extra structure, but you don't have to. You get equivalent categories with the following definitions. A graded vector space is a sequence $(V_n)_{n \in \mathbb{N}}$ of vector spaces -- the sum of elements in different grades is not taken to be well-defined. A morphism of graded vector spaces is a sequence $(f_n)_{n \in \mathbb{N}}$ of linear maps, and the tensor is given by $(V_n)_{n\in \mathbb{N}} \otimes (W_m)_{m \in \mathbb{N}} = (\oplus_{n+m=p} V_n \otimes W_m)_{p \in \mathbb{N}}$. A monoid in this monoidal category turns out to be exactly what you describe: a sequence of vector spaces $(A_n)_{n \in \mathbb{N}}$ with multiplication maps $V_n \otimes V_m \to V_{n+m}$ satisfying unit and associativity laws.

(There are actually two important ways to make this into a symmetric monoidal category: one symmetry is $\sigma (v_n \otimes w_m) = w_m \otimes v_n$ while the other is $\sigma(v_n \otimes w_m) = (-1)^{nm} w_m \otimes v_n$. Under the second symmetry, the exterior algebra is actually a commutative monoid!)

Anyway, the point is that the exterior algebra is a graded algebra, and the category of graded algebras can be defined in different ways. In some of these ways, we think of the sum of elements in different grades as being well-defined, while in others we don't. It's a matter of taste.

EDIT I can't resist pointing out that "taking the sums of elements of different grades to be well-defined" amounts to defining a functor $\mathsf{GrVect} \to \mathsf{Vect}$ from graded vector spaces to vector spaces, sending $(V_n)_{n \in \mathbb{N}} \mapsto \oplus_n V_n$. But this isn't even the only useful such functor -- another one would be $(V_n)_{n \in \mathbb{N}} \mapsto \Pi_n V_n$, which differs when there are infinitely many nonzero grades. Admittedly this functor doesn't play as well with the monoidal structure.

EDIT 2 The description I gave of $\mathsf{GrVect}$ essentially regards it as the functor category $[\mathbb{N}, \mathsf{Vect}]$ (where $\mathbb{N}$ is regarded as a discrete category). The monoidal product is given by Day convolution (where $\mathbb{N}$ is regarded as being monoidal under addition), a general way of producing monoidal structures on nice functor categories.

Nice question! Well, let's say you just have the $\Lambda^k(T(M))$ in isolation ($M$ is a manifold, $T(M)$ is its tangent space). What are they? Modules over the $C^n(M)$ functions ring? Nice, but how can we multiply one form with another form? We can't, we have modules not rings or algebras. Yet we would like to multiply a 1-form with another 1-form to build a 2-form, for example. OK, so we invent the exterior product $\wedge^{k,s}$. It is called exterior because it takes a k-form and a s-form and gives you back a (k+s)-form. It goes out of the original modules. We actually have a collection of these exterior products, one for each pair of $k,s$. So we have a collections of modules $\Lambda^k(T(M))$ and a collection of exterior products $\wedge^{k,s}$. What a mess. When faced with similar situations mathematicians just generalize things, to make them simpler. They did exactly the same thing with polynomials after all. Think about it: there is no internal multiplication between polynomials of order $k$.

So let's create a mathematical structure, $\Lambda(T(M))$, which contains all the $\Lambda^k(T(M))$. $\Lambda(T(M))$ is the direct sum of all the $\Lambda^k(T(M))$ There is now an internal multiplication which is called the wedge product $\wedge$, which subsumes all the $\wedge^{k,s}$. There is also an internal addition and scalar multiplication which subsume all the different additions and scalar multiplications in the $\Lambda^k(T(M))$. Actually the scalar multiplication is the same as the wedge product with $\Lambda^0(T(M))$. So what we have got is an algebra! Great, we went from a collection of structures and external operations to a single mathematical structure. Much cleaner. You can also create a single derivative internal operator which subsumes all the different exterior derivative operators acting on the individual $\Lambda^k(T(M))$.

But the most important construction using the full power of your new graded algebra $\Lambda(T(M))$ is the Ideal. This construction requires the freedom to add and multiply forms with different degrees. You cannot do it with modules, you need a ring at least.

I suggest you read "Applied exterior calculus" by Edelen to see the neat things you can do with $\Lambda(T(M))$

Now, do you still find it uncomfortable to sum a 1-form with a 2-form as in your $\omega$?

It boils down to psychology.

Think about it this way: when you were a kid they showed you an apple sitting next to another apple and they told you to describe this as 2 apples. In math you wrote $1+1=2$. You asked: can I do this with bananas too? Sure, you can do this with anything and the result is always $2$, but....if you have an apple and a banana you cannot sum them anymore. You can only sum apples with apples and bananas with bananas.

Bad,Bad,Bad teacher.

Fast forward to today and you do not know what to do with 1-form sitting next to a 2-form.

But suppose you had a better teacher. She shows you 3 and a half apples and 2 bananas next to each other on a table and ask the smart kid in your soul to describe this (the smart kid is there, just look for it).

The smart kid says: oh, as you told me yesterday, it's a vector in a 2D vector space over the rationals. I would describe it as $3.5 a + 2 b$ where $a$ and $b$ are base vectors "apple" and "banana".

So the teacher asks: what if I give you another half apple and take away a banana?

The kid replys: now I have $4a+b$, it's a vector space with basis a and b, so you still sum apples with apples and bananas with bananas, but if you have apples and bananas you just imagine you have a vector space with more dimensions instead of the usual 1D space of apples or bananas (which is equivalent to a scalar, that is a simple number).

The bad teacher thought that "apples" and "bananas" were numbers. They are in fact vectors in 1D vector spaces. When you sum them together, you are doing a "direct sum" of vector spaces.

Fast forward to today. You have a 1-form sitting next to a 2-form what do you do? Just add some milk and mix yourself a great mathematical smoothie :-)