The midpoints $M$ of $\overline{AB}$ and $N$ of $\overline{CD}$ lie on the common perpendicular bisector of those edges, and $f(X) \ge f(Q)$ for a point $Q$ on line $MN$

Rotate $D$ about line $MN$ into the plane $AMN:$ then $f(Q) = 2(AQ + D'Q) \ge 2AD',$ and the median length formula gives $AD'$

Let $M$ and $N$ be the midpoints of $\overline{AB}$ and $\overline{CD}.$ The medians from $C$ and from $D$ to $\overline{AB}$ are equal, since triangles $ABC$ and $BAD$ are congruent by $SSS;$ by the median length formula, $4MD^2 = 2 \cdot 28^2 + 2 \cdot 44^2$ $- 52^2$ $= 2736,$ so $MC^2 = MD^2 = 684.$ Likewise $NA = NB.$ Then $MN,$ as a median of the isosceles triangles $MCD$ and $NAB,$ is perpendicular to both $\overline{AB}$ and $\overline{CD},$ so the $180^\circ$ rotation about line $MN$ swaps $A \leftrightarrow B$ and $C \leftrightarrow D.$ Also $MN^2 = MD^2 - ND^2$ $= 684 - 26^2$ $= 8.$

For any point $X,$ let $X'$ be its image under this rotation, and let $Q$ be the midpoint of $\overline{XX'},$ which lies on line $MN.$ Then $BX = AX'$ and $CX = DX',$ so $$\begin{aligned} &f(X) = (AX + AX') \\ &{}+ (DX + DX') \\ &\ge 2AQ + 2DQ = f(Q), \end{aligned}$$ because a median of a triangle is at most half the sum of the two adjacent sides. So it suffices to minimize $f(Q) = 2(AQ + DQ)$ over points $Q$ on line $MN.$

Rotate $D$ about line $MN$ into the plane of $A$ and line $MN,$ on the opposite side of $MN$ from $A,$ landing at $D'$ with $ND' = ND = 26.$ For $Q$ on line $MN,$ $AQ + DQ = AQ + D'Q \ge AD',$ with equality where segment $\overline{AD'}$ crosses $MN.$ Since $AM \perp MN$ and $D'N \perp MN$ with $AM = 26$ and $ND' = 26,$ $$\begin{aligned} &AD'^2 = (AM + ND')^2 + MN^2 \\ &= 52^2 + 8 = 2712 \\ &= 4 \cdot 678. \end{aligned}$$ Hence the minimum of $f$ is $2AD' = 4\sqrt{678},$ and since $678 = 2 \cdot 3 \cdot 113$ is squarefree, $m + n = 4 + 678 = 682.$