Lets call the square $ABCD$ and say we would like to divide $AB$. On the line $BC$, pick a segment $e<\frac{AB}{n}$ and draw $n$ of these over $BC$, marking the points $F_j$ along the way. Connect $AF_n$ with the ruler and slide it parallel with such that every point we hit in $BC$ we mark a point in $AB$ where the ruler hits, dividing $AB$ into $n$ equal parts. Is this valid?

I don't think this is valid, unless this "sliding" is indeed allowed. Let's first show how to do bisect every side. Let the square be $ABCD$. First we can clearly construct the diagonals and the center of the square, call it $O.$ Select an arbitrary point $P$ on $AB.$ Let $CP$ meet $BD$ at $\epsilon$ and let $A \epsilon$ hit $BC$ at $Q.$ We have $PQ || AC.$ Let $PQ$ hit $BD$ at $\Gamma$ and let $C\Gamma, BD$ meet at $\gamma.$ Finally, let $B \gamma$ hit $AC$ at $T.$ We have $\frac{CT}{TA} = \frac13.$ Analogously, we can construct $U \in BD$ with $\frac{DU}{UB} = \frac13.$ Then construct $CU \cap DT.$ This is a point on the line connecting the midpoints of $AB$ and $CD.$ By symmetry we can also construct the reflection of this point over the center, and so connecting this point with its reflection gives us the midpoints of $AB, CD.$ Analogously, we can construct the midpoints of $BC, AD$, so we're done with $n = 2.$ Let's show how to do $n+1$, assuming that we're done with $n$. Let $A = A_0, A_1, \cdots, A_n = D$ be $n+1$ points dividing $AD$ into $n$ equal parts, and $B = B_0, B_1, \cdots, B_n = C$ be $n+1$ points dividing $BC$ into $n$ equal parts. Mark the points $C_i = B_{i-1}A_i \cap AC$, for all $1 \le i \le n.$ Also mark the points $D_i = A_{i-1}B_i \cap BD$ for all $1 \le i \le n.$ These points divide the two diagonals into $n+1$ equal parts. It remains only to connect the corresponding pairs of points to split $AD, BC$ into $n+1$ equal parts. Repeat with $AB, CD$ to finish. With the above two constructions, we can easily finish with induction. $\square$