Suppose that ideals $I$ and $J$ of $k[x_1,\dots,x_n]$ are given with $\{g_1,\dots,g_m\}$ and $\{f_1,\dots,f_n\}$ as their respective Grobner bases. Under what conditions is $\{g_1,\dots,g_m,f_1,\dots,f_n\}$ a Grobner basis for $I+J$ (with respect to the same monomial order)?
In particular, if $J=(x_n)$, what can be said about a Grobner basis of $I+(x_n)$?
The same question was asked here, however I'm unsure about the argument that if $f+g\in I+J$ then $\text{LT}(f+g)\in\text{LT}(I)\cup\text{LT}(J)$, where $\text{LT}(f)$ denotes the leading term of $f$ and $\text{LT}(I)$ is the set of leading terms of polynomials in $I$. Can it not be the case that if e.g. $\text{LT}(f)=-\text{LT}(g)$ then $\text{LT}(f+g)$ is a term that doesn't appear in $\text{LT}(I)\cup\text{LT}(J)$?
The observation in the last part of your question is correct. In general, given two polynomials $f$ and $g$ in some ideal $I$ of $k[x_1,\dots,x_n]$, it may happen that some combination $af+bg$ produces cancellation of the leading terms, yielding a polynomial in $I$ whose leading term is potentially not in $\langle \text{LT}(f),\text{LT}(g) \rangle$. Studying this cancellation phenomenon more precisely, one is to lead to the so-called $S$-polynomials (which are designed to produce cancellation) and further to Buchberger's criterion:
Here, the notation $f \rightarrow_G 0$ means that $f$ has a standard representation with respect to $G$, i.e. we can write \begin{equation*} f = h_1g_1 + \dots + h_sg_s, \end{equation*} where $\deg(f) \geq \deg(h_ig_i)$ whenever $h_ig_i \neq 0$. In practice, one way to obtain such a standard representation is to get remainder $0$ after dividing $f$ by $G$ (where the elements of $G$ are listed in some order). In the following, we will implicitly use the fact that whenever $G \subseteq H$, then $f \rightarrow_G 0$ implies that $f \rightarrow_H 0$.
Going back to your question, we know that $G = \{g_1,\dots,g_m,f_1,\dots,f_n\}$ is a basis for $I+J$, so Buchberger's criterion tells us that $G$ is a Gröbner basis if and only if \begin{equation*} S(g_i,g_j) \rightarrow_G 0,\quad S(f_k,f_\ell) \rightarrow_G 0 \quad\text{and}\quad S(g_i,f_k) \rightarrow_G 0 \end{equation*} for all $1 \leq i,j \leq m$ and all $1 \leq k,\ell \leq n$. But we already know this for $S(g_i,g_j)$ and $S(f_k,f_\ell)$ since $\{g_1,\dots,g_m\}$ and $\{f_1,\dots,f_n\}$ are Gröbner bases for $I$ and $J$, respectively. Therefore, only the $S(g_i,f_k)$ are left to consider.
In the particular case where $J = (x_n)$, the set $G = \{g_1,\dots,g_m,x_n\}$ is a Gröbner basis of $I+(x_n)$ if and only if $S(g_i,x_n) \rightarrow_G 0$ for all $1 \leq i \leq m$. For example, if $\{g_1,\dots,g_m\} \subseteq k[x_1,\dots,x_{n-1}]$, then the $S$-polynomials $S(g_i,x_n)$ are all divisible by $x_n$, so they leave remainder $0$ when divided by $G$ (this is easy to see if we list $G$ as $\{x_n,g_1,\dots,g_n\}$), hence $G$ is a Gröbner basis for $I +(x_n)$ in this case.
More generally, a sufficient condition for $S(g_i,f_k) \rightarrow_G 0$ to hold is that the leading terms of $g_i$ and $f_k$ are relatively prime, cf. this question or Proposition 4 in Chapter $2$, $\S9$ of the book Ideals, Varieties, and Algorithms by Cox, Little and O'Shea.