$\newcommand{\bu}{\boldsymbol{u}}$ $\newcommand{\bv}{\boldsymbol{v}}$ $\newcommand{\bP}{\mathbb{P}}$ $\newcommand{\bR}{\mathbb{R}}$
We define
$$F:S^{m-1}\times S^{m-1} \to \bR, \;\;F(\bu,\bv):=|\bu\cdot\bv|^2. $$ Note that the range of $F$ is $[0,1]$. Denote by $p(d\bv)$ the probability density of $\bv$
For every interval $[a,b]\subset (0,1)$ we have
$$ \bP[a\leq f\leq b] =\int_{S^{m-1}} \bP[a\leq F\leq b| \bv=\bv_0] p(d\bv_0) $$
$$= \int_{S^{m-1}} \Bigl(\;{\rm Area}\;\{ \sqrt{a}\leq |\bu\cdot\bv_0|\leq \sqrt{b}\}\;\Bigr)\; p(d \bv_0) $$
$$=2 \int_{S^{m-1}} \underbrace{ \Bigl(\;{\rm Area}\;\{ \sqrt{a}\leq \bu\cdot\bv_0\leq \sqrt{b}\}\;\Bigr)}_{=:I(a,b,\bv_0)}\; p(d \bv_0) $$
Due to the rotational symmetry, the integrand $I(a,b,\bv_0)$ is independent of $\bv_0$ so I will denote it by $I(a,b)$. Hence
$$\bP[a\leq f\leq b]= 2 I(a,b). $$
To compute $I(a,b)$ use the coarea formula exactly as in [Example 9.1.10 of these notes][1]. We have $\newcommand{\bsi}{\boldsymbol{\sigma}}$
$$ I(a,b) = \bsi_{m-2}\int_{\sqrt{a}}^{\sqrt{b}}(1-t^2)^{\frac{m-3}{2}} dt, $$
where $\bsi_k$ denotes the area of the unit $k$-dimensional sphere
$$\bsi_k=\frac{2\pi^{\frac{k+1}{2}}}{\Gamma\left(\frac{k+1}{2}\right)}. $$
If $\rho_F(x)$ denotes the probability density of $F$, then we deduce that
$$\rho_F(x)=2\frac{d}{dh}\Bigl|_{h=0} I(x,x+h)=\bsi_{m-2} x^{-1/2}(1-x^2)^{\frac{m-3}{2}}. $$
Thus the answer is independent of the distribution of $\bv$. [1]: http://www3.nd.edu/~lnicolae/Lectures.pdf