As noted in the comments, by the spherical symmetry, the distribution of the dot product in both parts of your question is the same that of $X\cdot(1,0,\dots,0)$. Moreover, the distribution of $X$ is the same as that of the random vector $$\frac{Z}{\sqrt{Z_1^2+\dots+Z_d^2}},$$ where $Z=(Z_1,\dots,Z_d)$ is a standard normal random vector. So, the distribution of the dot product in question is the same that of $$R:=\frac{Z_1}{\sqrt{Z_1^2+\dots+Z_d^2}}.$$ The distribution of $R$ is obviously symmetric, and the distribution of $R^2$ is the beta distribution with parameters $\frac12,\frac{d-1}2$. It follows that the probability density function (pdf) $f_R$ of $R$ is given by $$f_R(r)=\frac{\Gamma \left(\frac{d}{2}\right)}{\sqrt{\pi } \Gamma \left(\frac{d-1}{2}\right)}\,\left(1-r^2\right)^{\frac{d-3}{2}}\, 1\{|r|<1\},$$ and the dot product in question has the same pdf.

As noted in the comments, by the spherical symmetry, the distribution of the dot product in both parts of your question is the same that of $X\cdot(1,0,\dots,0)$. Moreover, the distribution of $X$ is the same as that of the random vector $$\frac{Z}{\sqrt{Z_1^2+\dots+Z_d^2}},$$ where $Z=(Z_1,\dots,Z_d)$ is a standard normal random vector. So, the distribution of the dot product in question is the same that of $$R:=\frac{Z_1}{\sqrt{Z_1^2+\dots+Z_d^2}}.$$ The distribution of $R$ is obviously symmetric, and the distribution of $R^2$ is the beta distribution with parameters $\frac12,\frac{d-1}2$. It follows that the probability density function (pdf) $f_R$ of $R$ is given by $$f_R(r)=\frac{\Gamma \left(\frac{d}{2}\right)}{\sqrt{\pi }\, \Gamma \left(\frac{d-1}{2}\right)}\,\left(1-r^2\right)^{\frac{d-3}{2}}\, 1\{|r|<1\},$$ and the dot product in question has the same pdf.