I want to define the tensor algebra and related algebras in a very formal way. I will illustrate how I tried to define algebras below.
Let $R$ be a commutative ring and $M$ be an $R$-module.
Consider the tensor algebra $T(M):=\bigoplus_{n\in \mathbb{N}} M^{\otimes n}$. Then there exists a canonical monomorphism $\pi_n:M^{\otimes n}\rightarrow T(M):n\mapsto \delta_n$ for each $n\in\mathbb{N}$. Then, let's define $T^n(M):=\pi_n(M^{\otimes n})$. So that $T^n(M)$ becomes really a subset of $T(M)$. And, for convenience, let's write $m_1\otimes...\otimes m_{n}$ to mean $\pi_n(m_1\otimes...\otimes m_n)$.
Now, let $I$ be the ideal of $T(M)$ generated by elements of the form $m_1\otimes m_2 - m_2\otimes m_1$. Then, take the quotient $R$-algebra of $T(M)$ by $I$ and denote it by $S(M)$ and call it the symmetric algebra on $M$.
Now, define $S^n(M):=\{x+I\in S(M): x\in T^n(M)\}$ and call it the $n$-th symmetric power of $M$.
Is my definition equivalent to the usual definition? That is, $S^k(M)$ is usually defined as $T^k(M)/I\cap T^k(M)$, but I hate this since this is not actually a subset of $S(M)$.. Is it okay to define $S^n(M)$ in my way?
The inclusion $T^k(M) \to T(M)$ induces a map $$ T^k(M)/I \cap T^k(M) \to T(M)/I = S(M) $$ which is obviously injective. Your definition of $S^k(M)$ is exactly the submodule of $S(M)$ which is the image of this map (since obviously $T^k(M) \to S(M)$ is trivial on $I \cap T^k(M)$). Therefore, the two definitions give canonically isomorphic modules, and it is unlikely to result in confusion if one uses one or the other.