$$p_i = \frac{\partial L}{\partial \dot{q}^i}$$ is the fundamental definition of generalized momentum. There is a very good reason for choosing it and that reason is the natural, canonical switch from the second derivative Euler-Lagrange equations of motion to the first derivative vector field form of Hamiltonian equations of motion. Indeed, the Euler-Lagrange equations are
$$\frac{d}{dt} \, \frac{\partial L}{\partial \dot{q}^i}\big(q,\dot{q}, t\big) = \frac{\partial L}{\partial {q}^i}\big(q,\dot{q}, t\big) $$ After setting $$p_i = \frac{\partial L}{\partial \dot{q}^i}\big(q,\dot{q}, t\big)$$ you can rewrite the Euler-Lagrange system as the double system
\begin{align}
p_i &= \frac{\partial L}{\partial \dot{q}^i}\big(q,\dot{q}, t\big)\\
\frac{d}{dt} p_i &= \frac{\partial L}{\partial {q}^i}\big(q,\dot{q}, t\big)
\end{align}
Solve the first set of equations with respect to $\dot{q} = f(q, p, t)$ and you get
\begin{align}
\dot{q}^i &= f^i(q, p, t)\\
\dot{p}_i &= \frac{\partial L}{\partial {q}^i}\big(q, \, f(q, p, t), \, t\big)
\end{align}
As it turns out, if you define the function
$$H(q, p, t) = \sum_{k}\, p_k \, \dot{q}^k - L(q, \dot{q}, t) = \sum_k\, p_k \, f^k(q,p,t) - L\big(q,\, f(q,p,t),\, t\big)$$ then
\begin{align}
\frac{\partial H}{\partial q^i}\big(q, p, t\big) &= \sum_k\, p_k \, \frac{\partial f^k}{\partial q^i}\big(q,p,t\big) - \frac{\partial L}{\partial q^i}\big(q,\, f(q,p,t),\, t\big) \\
& \,\,\,\,\,\, - \sum_k \, \frac{\partial L}{\partial \dot{q}^k}\big(q,\, f(q,p,t),\, t\big) \frac{\partial f^k}{\partial q^i}\big(q,\, p,\, t\big) = \\
&= \sum_k\, p_k \, \frac{\partial f^k}{\partial q^i}\big(q,p,t\big) - \frac{\partial L}{\partial q^i}\big(q,\, f(q,p,t),\, t\big)
- \sum_k \, p_k \frac{\partial f^k}{\partial q^i}\big(q,\, p,\, t\big) = \\
&= - \, \frac{\partial L}{\partial q^i}\big(q,\, f(q,p,t),\, t\big)
\end{align}
Hence $$\dot{p}_i = \frac{\partial L}{\partial q^i}\big(q,\, f(q,p,t),\, t\big) = - \,
\frac{\partial H}{\partial q^i}\big(q, p, t\big)$$
Analogously,
\begin{align}
\frac{\partial H}{\partial p_i}\big(q, p, t\big) &= f^i(q, p, t) + \sum_k\, p_k \, \frac{\partial f^k}{\partial p_i}\big(q,p,t\big)
- \sum_k \, \frac{\partial L}{\partial \dot{q}^k}\big(q,\, f(q,p,t),\, t\big) \frac{\partial f^k}{\partial p_i}\big(q,\, p,\, t\big) = \\
&= f^i(q, p, t) + \sum_k\, p_k \, \frac{\partial f^k}{\partial p_i}\big(q,p,t\big)
- \sum_k \, p_k \frac{\partial f^k}{\partial q^i}\big(q,\, p,\, t\big) = \\
&= - \, \frac{\partial L}{\partial p_i}\big(q,\, f(q,p,t),\, t\big) = \\
&= f^i(q, p, t)
\end{align}
Hence
$$\dot{q}^i = f^i\big(q, p , t\big) =
\frac{\partial H}{\partial p_i}\big(q, p, t\big)$$ Consequently the Euler-Lagrange system turns into the Hamiltonian system
\begin{align}
\dot{q}^i &=
\frac{\partial H}{\partial p_i}\big(q, p, t\big)\\
\dot{p}_i &= - \,
\frac{\partial H}{\partial q^i}\big(q, p, t\big)
\end{align}
The Hamiltonian form of the system has far reaching consequences.
Only in the case where the Lagrangian is of the form kinetic minus potential energy $$L\big(q, \dot{q}, t\big) = T\big(q, \dot{q}, t\big) \, - \, U\big( q, t\big)$$ with $T$ being the kinetic energy, we obtain
$$p_i = \frac{\partial L}{\partial \dot{q}^i} = \frac{\partial T}{\partial \dot{q}^i}$$ This is a special case.
However, in electrodynamics for example, you may encounter the Lagrangian of a charged particle with charge $c$ moving in an electromagnetic field. In this case, the Lagrangian is something like
$$L = T(q, \dot{q}) + \sum_{k=1}^3 c\,A_k\big(q, t\big) \, \dot{q}^k \, - \, c\,\phi\big(q, t\big) $$ where $A_k(q,t)$ is the magnetic vector potential and $\phi(q,t)$ is the electric scalar potential. As you can check here, the generalized momenta are
$$p_i = \frac{\partial L}{\partial \dot{q}^i} =
\frac{\partial T}{\partial \dot{q}^i}\big(q, \dot{q}\big) + c\, A_i\big(q, t\big) \neq
\frac{\partial T}{\partial \dot{q}^i}\big(q, \dot{q}\big)$$
