Remark on connections with the same geodesics:
I realize that I didn't respond to the OP's confusion about connections with the same geodesics vs. compatible with a metric $g$ but with torsion.
(I did respond in a comment that turned out to be wrong, so I deleted it. Hopefully, this will be better.)
First, about torsion (of a connection on TM). The torsion $T^\nabla$ of a (linear) connection on $TM$ is a section of the bundle $TM\otimes\Lambda^2(T^*M)$. Here is an (augmented) Fundamental Lemma of (pseudo-)Riemannian geometry:
Lemma 1: If $g$ is a (nondegenerate) pseudo-Riemannian metric on $M$ and $\tau$ is a section of $TM\otimes\Lambda^2(T^*M)$, then there is a unique linear connection $\nabla$ on $TM$ such that $\nabla g = 0$ and $T^\nabla = \tau$.
(The usual FLRG is the special case $\tau=0$.) Note that this $\nabla$ depends algebraically on $\tau$ and the $1$-jet of $g$. The proof of Lemma 1 is the usual linear algebra.
Second, if $\nabla$ and $\tilde\nabla$ are two linear connections on $TM$,
their difference is well-defined and is a section of $TM\otimes T^*M\otimes T^*M$. Specifically $\tilde\nabla - \nabla:TM\times TM\to TM$ has the property that, on vector fields $X$ and $Y$, we have
$$
\bigl(\tilde\nabla - \nabla\bigr)(X,Y) = \tilde\nabla_XY-\nabla_XY.
$$
Lemma 2: Two linear connections, $\nabla$ and $\tilde\nabla$ have the same geodesics (i.e., each curve $\gamma$ is a geodesic for one if and only if it is a geodesic for the other) if and only if $\tilde\nabla - \nabla$ is a section of the subbundle $TM\otimes\Lambda^2(T^*M)\subset TM\otimes T^*M\otimes T^*M$.
Proof: In local coordinates $x = (x^i)$, let $\Gamma^i_{jk}$ (respectively, $\tilde\Gamma^i_{jk}$) be the coefficients of $\nabla$0 (respectively, $\tilde\nabla$). Then
$$
\tilde\nabla-\nabla = (\tilde\Gamma^i_{jk}-\Gamma^i_{jk})\ \partial_i\otimes \mathrm{d}x^j\otimes\mathrm{d}x^k.
$$
Meanwhile, a curve $\gamma$ in the $x$-coordinates is a $\nabla$-geodesic
(respectively, a $\tilde\nabla$-geodesic) iff
$$
\ddot x^i + \Gamma^i_{jk}(x)\,\dot x^j\dot x^k = 0\qquad
(\text{respectively},\ \ddot x^i + \tilde\Gamma^i_{jk}(x)\,\dot x^j\dot x^k = 0).
$$
These are the same equations iff $(\tilde\Gamma^i_{jk}(x)-\Gamma^i_{jk}(x))\,y^jy^k\equiv0$ for all $y^i$, i.e., iff
$$
{\tilde\nabla}-\nabla = \tfrac12({\tilde\Gamma}^i_{jk}-\Gamma^i_{jk})\ \partial_i\otimes \mathrm{d}x^j\wedge\mathrm{d}x^k.\quad \square
$$
Finally, we examine when two $g$-compatible connections have the same geodesics:
Lemma 3: If $g$ is a nondegenerate (pseudo-)Riemannian metric, and $\nabla$ and $\tilde\nabla$ are linear connections on $TM$ that satisfy $\nabla g = \tilde\nabla g = 0$, then they have the same geodesics if and only if the expression
$$
\phi(X,Y,Z) = g\bigl( X,(\tilde\nabla{-}\nabla)(Y,Z)\bigr)
$$
is skew-symmetric in $X$, $Y$, and $Z$.
Proof: $\nabla g = \tilde\nabla g = 0$ implies $\phi(X,Y,Z)+\phi(Z,Y,X)=0$, while they have the same geodesics if and only if $\phi(X,Y,Z)+\phi(X,Z,Y)=0$.
Corollary: If $g$ is a nondegenerate (pseudo-)Riemannian metric, then the space of linear connections $\nabla$ on $TM$ that satisfy $\nabla g = 0$ and have the same geodesics as $\nabla^g$, the Levi-Civita connection of $g$, is a vector space naturally isomorphic to $\Omega^3(M)$, the space of $3$-forms on $M$.