According to equation (6) on the first page of some lecture notes online, the above equation is used to prove the virial theorem. For rectangular coordinates, the relation
$$ 2T~=~\sum_i p_{i}\dot{q}^{i} $$
is obvious. How would I show it holds for generalized coordinates $q^{i}$?
Ok so the answer is more straightforward and tedious than I expected.
In rectangular coordinates, the kinetic energy $T$ of $n$ particles is:
$$ T=\frac{1}{2}\sum_{\alpha=1}^{n}\sum_{i=1}^{3}m_{\alpha}\dot{x}_{\alpha,i}^{2} $$
where $x_{\alpha,i}$ is a function of generalized coordinates:
$$ \begin{eqnarray} x_{\alpha,i} &=& x_{\alpha,i}(q_{j},t),\hspace{12pt}j=1,2,...,s\\ \dot{x}_{\alpha,i} &=& \sum_{j=1}^{s}\frac{\partial x_{\alpha,i}}{\partial q_{j}}\dot{q}_{j}+\frac{\partial x_{\alpha,i}}{\partial t} \end{eqnarray} $$
So $\dot{x}_{\alpha,i}^{2}$ becomes
$$ \dot{x}_{\alpha,i}^{2}=\sum_{j,k}\frac{\partial x_{\alpha,i}}{\partial q_{j}}\frac{\partial x_{\alpha,i}}{\partial q_{k}}\dot{q}_{j}\dot{q}_{k} + 2\sum_{j}\frac{\partial x_{\alpha,i}}{\partial q_{j}}\frac{\partial x_{\alpha,i}}{\partial t}\dot{q}_{j} + \left(\frac{\partial x_{\alpha,i}}{\partial t}\right)^{2} $$
The total kinetic energy is:
$$ T=\sum_{\alpha}\sum_{i,j,k}\frac{1}{2}m_{\alpha}\frac{\partial x_{\alpha,i}}{\partial q_{j}}\frac{\partial x_{\alpha,i}}{\partial q_{k}}\dot{q}_{j}\dot{q}_{k} + \sum_{\alpha}\sum_{i,j}m_{\alpha}\frac{\partial x_{\alpha,i}}{\partial q_{j}}\frac{\partial x_{\alpha,i}}{\partial t}\dot{q}_{j} + \sum_{\alpha}\sum_{i}\frac{1}{2}m_{\alpha}\left(\frac{\partial x_{\alpha,i}}{\partial t}\right)^{2} $$
If $x_{\alpha, i}$ has no explicit time dependence (as is the case if the system is scleronomic), then $\frac{\partial x_{\alpha ,i}}{\partial t}=0$, so the total kinetic energy becomes:
$$ \begin{eqnarray} T &=& \sum_{\alpha}\sum_{i,j,k}\frac{1}{2}m_{\alpha}\frac{\partial x_{\alpha,i}}{\partial q_{j}}\frac{\partial x_{\alpha,i}}{\partial q_{k}}\dot{q}_{j}\dot{q}_{k}\\ &=& \sum_{j,k}\left(\sum_{\alpha}^{n}\sum_{i}^{3}\frac{1}{2}m_{\alpha}\frac{\partial x_{\alpha,i}}{\partial q_{j}}\frac{\partial x_{\alpha,i}}{\partial q_{k}}\right)\dot{q}_{j}\dot{q}_{k}\\ T &=& \sum_{j,k}a_{jk}\dot{q}_{j}\dot{q}_{k} \hspace{12pt} \square \end{eqnarray} $$
Taking the partial derivative of $T$ with respect to $\dot{q}_{l}$ gives us:
$$ \begin{eqnarray} \frac{\partial T}{\partial \dot{q}_{l}} &=& \sum_{k}a_{lk}\dot{q}_{k} + \sum_{j}a_{jl}\dot{q}_{j}\\ \dot{q}_l\frac{\partial T}{\partial \dot{q}_{l}} &=& \sum_{k}a_{lk}\dot{q}_{k}\dot{q}_{l} + \sum_{j}a_{jl}\dot{q}_{j}\dot{q}_{l}\\ \end{eqnarray} $$
Summing over all $l$ gives us:
$$ \sum_{l}\dot{q}_l\frac{\partial T}{\partial \dot{q}_{l}} = \sum_{k,l}a_{lk}\dot{q}_{k}\dot{q}_{l} + \sum_{j,l}a_{jl}\dot{q}_{j}\dot{q}_{l} $$
All indices are dummy indices so:
$$ \sum_{l}\dot{q}_l\frac{\partial T}{\partial \dot{q}_{l}} = 2\sum_{j,k}a_{jk}\dot{q}_{j}\dot{q}_{k} = 2T \hspace{12pt} \square $$
If the potential doesn't depend on the generalized velocities, then:
$$ \begin{eqnarray} \sum_{l}\dot{q}_l p_{l} = \sum_{l}\dot{q}_l\frac{\partial L}{\partial \dot{q}_{l}} &=& \sum_{l}\dot{q}_l\frac{\partial (T-U)}{\partial \dot{q}_{l}}\\ &=& \sum_{l}\dot{q}_l\frac{\partial T}{\partial \dot{q}_{l}}\\ \sum_{l}\dot{q}_l p_{l} &=& 2T \hspace{12pt} \blacksquare \end{eqnarray} $$
So I guess the equation I originally asked does hold in general, given the two assumptions:
Your answer looks complicated and I feel it misses the point. The fact that $2T=\sum_l p_l \dot{q}_l$ is fundamentally due to Euler's homogeneous function theorem. This states that $$ \text{if }f(\alpha \mathbf{x})=\alpha^k f( \mathbf{x})\text{, then } \mathbf{x}\cdot\nabla f(\mathbf{x})=kf(\mathbf{x}). $$ When stated like that, I would even hesitate to add a proof - simply differentiate with respect to $\alpha$ and set it to $1$.
The theorem holds for Lagrangians that depend quadratically and homogeneously in the velocities. This means specifically that $L(q_l,\dot q_l)=T(\dot q_l;q_l)-V(q_l)$, and $T(\alpha \dot q_l;q_l)=\alpha^2 T(\dot q_l;q_l)$. (I use the notation $T(\dot q_l;q_l)$ to emphasize that the $q_l$ dependence is as parameters, with $T$ being fundamentally a homogeneous quadratic function of the $\dot q_l$.) The homogeneous function theorem then states that $$2T(\dot q_l;q_l)=\sum_l\dot q_l \frac{\partial T}{\partial \dot q_l}.$$ Since $V$ is independent of the $\dot q_l$, each partial derivative equals the corresponding momentum $p_l=\frac{\partial L}{\partial \dot q_l}$.
Note in particular that your first assumption (that the functional dependence of $T$ on the $\dot q_l$ be time-independent) is not necessary.
The equation $2T=\sum_{i=1}^n p_i\dot{q}^i$ holds both in Lagrangian and Hamiltonian formalism for large classes of systems.
In Lagrangian formalism, it holds for Lagrangians of the form $L(q,\dot{q},t)=T(q,\dot{q},t)-V(q,t)$, where the kinetic energy is of the form $T=\frac{1}{2}\sum_{i,j=1}^n \dot{q}^i m_{ij}(q,t)\dot{q}^j$. Now use the Lagrangian definition of momentum $p_i = \frac{\partial L}{\partial \dot{q}^i}$.
In Hamiltonian formalism, it holds for Hamiltonians of the form $H(q,p,t)=T(q,p,t)+V(q,t)$, where the kinetic energy is of the form $T=\frac{1}{2}\sum_{i,j=1}^n p_i m^{ij}(q,t)p_j$. Now use Hamilton's equations $\dot{q}^i = \frac{\partial H}{\partial p_i}$.
In both cases, we are secretly using the Euler homogeneous vector field $\sum_{i=1}^n\dot{q}^i\frac{\partial}{\partial \dot{q}^i}$, which counts the number of $\dot{q}$s, as Emilio Pisanty also points out in his answer.
