Do $S^n$ and $\mathbb{R}P^n$ admit the same number of linearly independent vector fields?

The quantity $i(M)$ is called the span of $M$.

First of all, let's establish the inequality $i(S^n) \geq i(\mathbb{RP}^n)$.

Let $M$ be a connected smooth manifold and $\Gamma$ a discrete group which acts smoothly, freely, and properly. Then $M/\Gamma$ can be equipped with a unique smooth manifold structure such that the projection $\pi : M \to M/\Gamma$ is a smooth covering map. In particular, $d\pi : TM \to T(M/\Gamma)$ is a vector bundle isomorphism covering $\pi$, so $\pi^*T(M/\Gamma) \cong TM$.

A manifold admits $k$ linearly independent vector fields if and only if its tangent bundle is isomorphic to a direct sum of two vector bundles, one of which is trivial of rank $k$. Suppose then that $M/\Gamma$ admits $k$ linearly independent vector fields, that is $T(M/\Gamma) \cong F\oplus\varepsilon^k_{M/\Gamma}$. Then we have

$$TM \cong \pi^*T(M/\Gamma) \cong \pi^*(F\oplus\varepsilon^k_{M/\Gamma}) \cong \pi^*F\oplus\pi^*\varepsilon^k_{M/\Gamma} \cong \pi^*F\oplus\varepsilon^k_M.$$

Therefore $M$ also admits $k$ linearly independent vector fields, so $i(M) \geq i(M/\Gamma)$; that is, the span of a manifold cannot decrease when passing to covers. In particular, for $M = S^n$ and $\Gamma = \mathbb{Z}_2$ acting by the antipodal map, we see that $i(S^n) \geq i(\mathbb{RP}^n)$ as you suspected.

For the other direction, note that in chapter I, section $7$ of Spin Geometry by Lawson and Michelsohn, they construct $k$ linearly independent vector fields $V_1, \dots, V_k$ on $S^n$ when $\mathbb{R}^{n+1}$ is a $Cl_k$-module. Moreover, these vectors fields satisfy $V_i(-x) = -V_i(x)$ and hence descend to $k$ linearly independent vector fields on $\mathbb{RP}^n$.

They then show how to determine the largest $k$ such that $\mathbb{R}^{n+1}$ is a $Cl_k$-module, i.e. the largest number of linearly independent vector fields on $S^n$ (and hence $\mathbb{RP}^n$) which can be constructed this way. They show it is equal to $\rho(n+1) - 1$ where $\rho(n)$ denotes the $n^{\text{th}}$ Radon-Hurwitz number which is defined as follows: if $n = 2^{4a + b}c$ where $a, b, c$ are non-negative integers, $0 \leq b \leq 3$ and $c$ is odd, then $\rho(n) = 2^b + 8a$. Therefore $i(S^n), i(\mathbb{RP}^n) \geq \rho(n+1) - 1$.

The final piece of the puzzle is the result by Adams which states that there are at most $\rho(n + 1) - 1$ linearly independent vector fields on $S^n$ so $i(S^n) = \rho(n+1) - 1$. But we also have $\rho(n + 1) - 1 = i(S^n) \geq i (\mathbb{RP}^n) \geq \rho(n+1) - 1$, so $i(\mathbb{RP}^n) = \rho(n+1) - 1$. Therefore, $i(S^n) = i(\mathbb{RP}^n)$.

Adams' result is very difficult, so it may seem that there is a more elementary argument to show that $i(S^n) = i(\mathbb{RP}^n)$. Note however that the equality $i(M) = i(M/\Gamma)$ does not hold in general, so an argument specific to this case was needed. For example, the two-dimensional torus is a double cover of the Klein bottle, and while the torus admits two linearly independent vector fields (i.e. it is parallelisable), the Klein bottle only admits one (it admits at least one as $\chi(K) = 0$, but it can't admit two as it is non-orientable). That is, $i(S^1\times S^1) = 2$ and $i(K) = 1$, so $i(S^1\times S^1) \neq i(K)$.