If $F_1$ and $F_2$ are free modules over a ring with the invariant dimension property, then $\text{rank} (F_1 \oplus F_2) = \text{rank} F_1 + \text{rank} F_2$.
I have written proof of this exercise below. My intention to write proof was to show idea used in proof is quite common in finding rank (dimension) of external direct sum-type problem.
Let $X$ and $Y$ be basis of $R$-module $F_1$ and $F_2$, respectively. We claim $$B=\{(x,0)\mid x\in X\} \cup \{(0,y)\mid y\in Y\}$$ is basis of $F_1\oplus F_2$.
Let $(u,v)\in F_1\oplus F_2$. Then $$\begin{align}(u,v) &= (\sum_{i=1}^na_ix_i, \sum_{i=1}^mb_iy_i)\\ &= \sum_{i=1}^na_i(x_i,0) + \sum_{i=1}^mb_i(0,y_i) \\ &\in \text{span} (B). \end{align}$$
If $\sum_{i=1}^na_i(x_i,0) + \sum_{i=1}^mb_i(0,y_i)=(0,0)$, then $\sum_{i=1}^na_i x_i=0$ and $\sum_{i=1}^mb_i y_i=0$. Since $X$ and $Y$ are linearly independent, $a_i=b_i=0$. Thus $B$ is linearly independent and basis of $F_1\oplus F_2$.
Since $R$ has dimension invariant property, we have $$\begin{align} \text{rank}(F_1\oplus F_2) &= |B| \\ &= | \{(x,0)\mid x\in X\} \cup \{(0,y)\mid y\in Y\}| \\ &= | \{(x,0)\mid x\in X\} | + | \{(0,y)\mid y\in Y\} | \\ &= |X|+|Y| \\ &= \text{rank} F_1 +\text{rank} F_2. \end{align}$$ Hence $\text{rank}(F_1\oplus F_2)= \text{rank} F_1 +\text{rank} F_2$.