Let's work in $R$-mod for the remainder of this question. I know I'm probably missing something super basic, but here goes. Let $F$ be the functor $- \otimes_R M$ for some $R$-module $M$. Let's use the definition (see https://en.wikipedia.org/wiki/Exact_functor) that a functor $F$ is right-exact iff $A \to B \to C \to 0$ is exact implies $F(A) \to F(B)\to F(C) \to 0$ is exact; and that a functor $F$ is exact iff $A\to B\to C$ exact implies $F(A)\to F(B)\to F(C)$ exact.
Then, suppose we have that $F$ is right-exact, and we are given $A \xrightarrow{f} B\xrightarrow{g} C$ exact. Then, the sequence $A \xrightarrow{f} B \xrightarrow{g} \text{im}(g) \to 0$ is an exact sequence. By right-exactness of $F$, we have $A \otimes_R M \xrightarrow{f \otimes \text{id}} B \otimes_R M \xrightarrow{g \otimes \text{id}} \text{im}(g) \otimes_R M \to 0$ is exact. But $\text{im}(g) \otimes_R M$ is a subset of $C \otimes_R M$, so we can think of $A \otimes_R M \xrightarrow{f \otimes \text{id}} B \otimes_R M \xrightarrow{g \otimes \text{id}} \text{im}(g) \otimes_R M$ as $A \otimes_R M \xrightarrow{f \otimes \text{id}} B \otimes_R M \xrightarrow{g \otimes \text{id}} C \otimes_R M$. In either case, the kernel of $g \otimes \text{id}$ equals the image of $f \otimes \text{id}$, so indeed $F$ satisfies the definition of exact.
Where did I go wrong?
Just answering my own question from the discussion in the comments; user Zhen Lin points out that tensoring doesn't preserve monomorphisms, i.e. "things that used to be submodules may no longer be submodules". An explicit example is $2 \mathbb Z \hookrightarrow \mathbb Z$, after tensoring with $\mathbb F_2$ (over $\mathbb Z$), becomes the zero map $\mathbb F_2 \to \mathbb F_2$ (the tensor products equal $\mathbb F_2$ because of the fact that for any $A$-module $M$, $A \otimes_A M \simeq M$, so $\mathbb Z \simeq 2 \mathbb Z \implies \mathbb Z \otimes_{\mathbb Z} \mathbb F_2 \simeq 2\mathbb Z \otimes_{\mathbb Z} \mathbb F_2 \simeq \mathbb F_2$. Warning, because I made this mistake: it's not true that the basis elements of $2 \mathbb Z \otimes \mathbb F_2$ are $(2n \otimes x) = (n \otimes 2x) = (n\otimes 0) = 0_{2 \mathbb Z \otimes \mathbb F_2}$, because $(n\otimes 2x)$ may not be in $2 \mathbb Z \otimes \mathbb F_2$! E.g. $(1 \otimes 1) \not\in 2\mathbb Z\otimes \mathbb F_2$).
So, although both tensor products may be generated by the same symbols, there may be some relations (like in the example above, the fact that $\mathbb F_2$ makes the multiplication-by-2 map the zero-map) that make the original injective function no longer injective.