Can we often only consider homogeneous elements of exterior or tensor algebras because their products preserve homogeneity? Background: I'm just an innocent physicist with very little formal training in algebra. So I think of things in very naive, concrete (as opposed to abstract), and non-rigorous ways. I probably won't understand an answer that contains much jargon. I think I understand the situation to a level of rigor that I find satisfactory, but I just want to verify my understanding.
I've sometimes heard the heuristic statement that

One big difference that distinguishes the geometric and Clifford algebras from the exterior and tensor algebras is that for the geometric and Clifford algebras, you can add together $k$-vectors with different grades $k$, but for the exterior and tensor algebras, you can't.

Now strictly speaking, I don't think this can be true. Formally, these are all algebras, so by definition you can always add together any two elements to get another element.
More concretely, (I believe that) all of these algebras are quotient algebras of the fundamental tensor algebra
$$T(V) := \bigoplus_{k=0}^\infty T^kV := \bigoplus_{k=0}^\infty V^{\otimes k} := \bigoplus_{k=0}^\infty \bigotimes_{i=1}^k V_i.$$
So strictly speaking, any element of any of these algebras can be a direct sum of nonzero $k$-vectors of different grades $k$. (A $k$-vector is an element of the algebra that can be written in the form $0_0 \oplus 0_1 \oplus \dots \oplus 0_{k-1} \oplus v_k \oplus 0_{k+1} \oplus \dots$.)
But if I understand directly, the key difference between Clifford/geometric algebras and exterior/tensor algebras is that for the tensor and exterior algebras, the product of a $k$-vector and an $l$-vector is always  $(k+l)$-vector. But for the Clifford and geometric algebras, the product of a $k$-vector and an $l$-vector will not necessarily be homogenous within any grade.
This means that, while we formally can add multivectors with different grades in the tensor and geometric algebras, in practice we can often "get away with" only considering homogeneous $k$-vectors and never adding across grades. Addition of multivectors obviously doesn't mix across different grades (by the definition of the direct sum), and since the product of homogeneous multivectors is homogeneous, we get a nice closed (under the product operation) subset of the algebra. So given any kind of non-homogenous equation, we can usually separate out each homogeneous subspace and consider the equation as several independent equations, without significant loss of generality.
But for geometric and Clifford algebras, this doesn't work: the product operation doesn't preserve homogeneity, so you need to consider fully general non-homogeneous elements of mixed grade.
Is that the right way to think about things? Obviously this isn't a very precise question, but I'd like to check that there's nothing above that really rings wrongly to the experts.
 A: 
Now strictly speaking, I don't think this can be true. Formally, these are all algebras, so by definition you can always add together any two elements to get another element.

This is correct.

But if I understand directly, the key difference between Clifford/geometric algebras and exterior/tensor algebras is that for the tensor and exterior algebras, the product of a -vector and an -vector is always (+)-vector. But for the Clifford and geometric algebras, the product of a -vector and an -vector will not necessarily be homogenous within any grade.

This is also correct.

This means that, while we formally can add multivectors with different grades in the tensor and geometric algebras, in practice we can often "get away with" only considering homogeneous -vectors and never adding across grades. Addition of multivectors obviously doesn't mix across different grades (by the definition of the direct sum), and since the product of homogeneous multivectors is homogeneous, we get a nice closed (under the product operation) subset of the algebra. So given any kind of non-homogenous equation, we can usually separate out each homogeneous subspace and consider the equation as several independent equations, without significant loss of generality.

This is also correct. Formally, the exterior and tensor algebras are ($\mathbb{Z}_{\ge 0}$-)graded algebras, meaning exactly that every element is a sum of homogeneous elements of various degrees and that a product of a homogeneous element of degree $k$ and a homogeneous element of degree $\ell$ is a homogeneous element of degree $k + \ell$. Clifford algebras do not have this property, but they are $\mathbb{Z}_2$-graded; there is still a "$\bmod 2$" degree that allows you to write every element as a sum of "even" and "odd" components, and the product of an even element with an even element is even, etc.
