Addendum: if you know Lagrangian mechanics, there the generalised momentum is defined to be $$\frac{\partial L(q,\dot{q}, t)} {\partial \dot{q}}$$
because this is the thing that is conserved if one of the coordinates is cyclic. This is clearly a linear function on the generalised velocities, so you can identify it with a covector.
For a free particle $L = \frac{1}{2} m \dot{q} ^2$ so that Lagrange's equation implies $$\frac{d}{dt}\frac{\partial L(q,\dot{q}, t)} {\partial \dot{q}} = 0 \implies m\dot{q} = const$$
Note that this is no longer a statement about $p=mv$ itself, but about a linear function of it. In other conditions there may be nothing interesting to say about $m\dot{q}$, but if the Lagrangian doesn't depend on $q$, $\partial L / \partial \dot{q}$ will still be conserved.
Short answer: for a coordinate system $(q^1,...q^n)$ on a manifold $M$ we let the generalised momenta $(p_1,...,p_n)$ be a basis for the contangent space which acts on $\lambda \in \pi^{-1}(M) \subset T^*M$ by $p_i(\lambda) = \lambda(\frac{\partial}{\partial q^i})$ where $\pi: T^*M \rightarrow M$ is the projection map . This gives the same results when $M$ is a vanilla vector space even though here momentum is not quite $p=mv$.
The underlying reason for this is that in Hamiltonian mechanics, the physics actually happens in the cotagent bundle, the 2n dimensional manifold parametrized by $(q^1 \circ \pi, ... , q^n \circ \pi, p_1, ..., p_n)$.
This formalism is motivated by the somewhat symmetric hole played by the $q^i$ and the $p_i$ in Hamilton's equation, so that we eventually forget about the base manifold and actually consider arbitrary 2n dimensional manifolds equipped with a anti-symmetrical non degenerate differential form (the symplectic form) which distinguishes the position from the momenta.
The topic is too big to explain in detail here, but looking at mechanics in this way gives you many deep results relatively easily. For instance, conservation on the symplectic form under motions implies conservation of volume of phase space. Also, due to the similarities between Hamilton's equations and the Cauchy Riemann equations, complex analysis methods can give some insight. This is the field of pseudoholomorphic curves.
For an introduction see the last chapters of Spivak's Physics for Mathematicians.
Let $T(q, \dot q)$ denote the kinetic energy of the system at configuration $q$ and velocity $\dot q$. On the tangent plane of some fixed point $q_0$, the kinetic energy induces a norm $\frac{1}{2}\|\dot q\|^2 = T(q_0, \dot q)$ on tangent vectors, which can be extended to an inner product via the polarization identity, and from there to a kinetic energy metric $g_T$ on the entire tangent bundle $TQ$. In this way kinetic energy gives $Q$ its geometric structure.
Momenta live on the cotangent space because they are dual to velocities, in the sense of the musical isomorphisms: a configurational velocity $\dot q$ and its corresponding momentum $p = \dot q^{\flat}$ satisfy, for any other tangent vector $v$,
$$pv = \langle \dot q, v\rangle_{g_T}$$
and in particular, $p\dot q = 2T(q,\dot q).$
Best Answer
The 1-form $p(x,x')$ is linear in the velocities in the following sense: sticking with your notation, the $i$th component of the one form is $$ p_i(x,x') = \frac{\partial L}{\partial {x'}^i}(x,x'). $$ The 1-form $p(x,x')$ acts linearly on velocities $v\in T_xM$ by $$ p(x,x')(v) = p_i(x,x')v^i = \frac{\partial L}{\partial {x'}^i}(x,x')v^i. $$ In other words, the linear dependence comes not through the $x'$ as you seem to believe, but through the $v$.