Choosing a left-invariant metric does indeed give you many one-parameter subgroups of isometries from which you can pick out Killing fields. However, these Killing fields will be right-invariant vector fields on $G$ (i.e. $X(g) = dR_g(X(e))$) and in general, these are not left-invariant. You can find a one-parameter subgroup of isometries by defining $\phi_s(g) = L_{exp(sv)}(g)$ for $s$ sufficiently small and $g \in G$ for some fixed $v \in \frak{g}.$ The vector field associated to this flow, $X$, can be found by computing
\begin{align} X(g) = X (\phi_0(g)) &= \frac{d}{ds}\mid_{s=0} exp(sv)\cdot g \\
&= \frac{d}{ds}\mid_{s=0} R_g (exp(sv))\\
&= dR_g( X(e)) \end{align}
So, left-invariant vectors are not necessarily Killing fields, and as you stated, one-parameter subgroups are not geodesics. The best way to see this is to take your favorite Lie group, $G$, and give it some left-invariant metric, $\langle \cdot, \cdot \rangle$ and study the one-parameter subgroups and left-invariant vector fields. Note, a left-invariant metric is a name for Riemannian metric you described above (i.e. choose an inner product $\langle \cdot, \cdot \rangle_{e_G}$ and use you left translations $L_g: G \rightarrow G$ that send $h\in G \mapsto gh \in G$ to define that inner product at each other tangent space of your manifold). I should also remark that even though all inner-product spaces are isometric, left-invariant metrics do not produce isometric Riemannian manifolds. The choice of initial inner product can dramatically alter the geometry.
My favorite example of a Lie group is the 3-dimensional $SL(2,\mathbf{R})$. This has a Lie algebra $\mathfrak{sl}(2,\mathbf{R})$ which has a basis $B =\left \{e,f,h \right \}$ with bracket relations:
\begin{equation} [h,e]=2e, \quad [h,f]=-2f, \quad [e,f]=h\end{equation}
Let's just choose an inner product on $\frak{sl}(2,\mathbf{R})$ so that $B$ is an orthonormal basis. The left-invariant vector field $X_e$ corresponding to $e$ is not a Killing field for this metric.
Claim: $X_e$ is not Killing.
Proof: A vector field $X$ is Killing iff $X$ sastifies the Killing equation
\begin{equation} \langle \nabla_Y X ,Z \rangle + \langle \nabla_Z X, Y \rangle =0 \quad \quad \text{for all} \quad Y,Z \in \mathfrak{X}(M) \end{equation}
(see: page 82 of Riemannian Geometry by do Carmo)
Well, if this is true, then that equation holds if $Y,Z$ are also left-invariant vector fields on $SL(2,\mathbf{R})$ (note, not every smooth vector field is left-invariant). Let's take $Y = X_h$ and $Z=X_f$ i.e. the left-invariant vector fields corresponding to the Lie algebra elements $h$ and $f$. One can compute that
\begin{equation} \langle \nabla_{X_h} X_e ,X_f \rangle + \langle \nabla_{X_f} X_e, X_h \rangle = \langle h, h \rangle_0 = 1 \end{equation}
where $\langle \cdot, \cdot \rangle_0$ is the inner product on $\frak{sl}(2,\mathbf{R}).$ $\Box$
Now, let's study one-parameter subgroups. Let $v \in \frak{sl}(2,\mathbf{R})$ and consider the one-parameter subgroup associated to $e$ i.e. $f_e(t) = exp(t e)$ for sufficiently small $t$.
Claim: $f_e(t)$ is not geodesic
Proof: Suppose by way of contradiction that this is a geodesic. Then, $\nabla_{X_e} X_e\equiv 0$ along the points on that curve. However, using the fact that we have a left-invariant metric, we can compute for all $v \in \frak{sl}(2,\mathbf{R})$,
\begin{equation} \langle X_v , \nabla_{X_e} X_e \rangle = \langle e, [v, e] \rangle_0 =0 \end{equation}
where $X_v$ is associated left-invariant vector field to $v$. This is clearly false for $v=h$. In general, that inner product vanishing for all $v$ is equivalent to the adjoint map of $ad_e: v \mapsto [e,v] $ is so that $ad^t_e(e) = 0$ where $(\cdot)^t$ denotes the transpose with respect to $\langle \cdot, \cdot \rangle_0$. If you write out the matrix associated to $ad_e$ written with respect to the orthonormal basis $B$, you can see this explicitly. $\Box$
Unfortunately, invariant Lie group geometry is not as simple as one might wish. However, there is a special class of Lie groups that do admit left-invariant metrics with geodesics corresponding to one-parameter subgroups and all left-invariant vector fields are Killing fields. These are Lie groups that admit a bi-invariant metric i.e. the metric can be equivalently defined using left-translations or right-translations. Any Abelian Lie group certainly have bi-invariant metrics. If one knows about the Killing form of a Lie algebra, you can see that compact semi-simple Lie groups also admit bi-invariant metrics. It turns out these are the only Lie groups that admit such metrics (Curvatures of left invariant metrics on lie groups, John Milnor, 1976--very readable!). You might look at exercise 3, Chapter 3 of do Carmo's book to see why bi-invariant metrics have these properties.
In light of these facts/computations, you might think of left-invariant metrics that don't have these properties as somehow having left translations and right translations that in some way do not mesh well geometrically.
Hint, but not a complete solution
For a moment, forget about Lie groups. Suppose that $f : M \to N$ is a diffeomorphism of manifolds. Then if you have a tangent vector $v$ at $m \in M$, you can compute $df(m)[v]$ to get a tangent vector at $n = f(m)$. This gives you, for each possible point of $N$, a tangent vector, and that's all that a vector field is. To be more explicit, suppose that we write $h = f^{-1}$, and pick any point $n \in N$. Let $X$ denote a vector field on $M$. Then our new vector field $Y$ on $N$ is defined by
$$
Y(n) = df(h(n))[X(h(n)],
$$
i.e., we compute the point $m = h(n)$ that's sent to $n$; look at the original vector field there (i.e., $X(h(n))$), and push it forward by the differential of $f$.
In the case of your Lie group, the manifolds $M$ and $N$ are both $G$, and the map $f$ is $L_q$, and the map $h$ is $L_{g^{-1}}$, but everything else still applies.
BTW, this whole construction of the "pushforward" of a vector field fails in general if $f$ is not a diffeomorphism, for $f^{-1}(n)$ might consist of multiple points, and pushing forward the vectors from all those points might lead to inconsistent results.
For your second question, I think you need to
Write down the definition of what it means for $X$ or $Y$ to be left-invariant. It's a little different from the definition of what it means for $\nabla$ to be left-invariant.
Let $Z = \nabla_X Y$. Use your definition from step 1 to say what it means for $Z$ to be left-invariant. That'll involve evaluating $Z(gh)$ and $Z(h)$ and comparing. To express those two things in a way that makes them comparable, you'll use d $L_g \nabla_X Y = \nabla_{d L_g X}\hspace{0.5mm} d L_g Y$,
probably evaluated at the point $h$, so you should write that out as well. And then just simplify with some algebra. I'm pretty sure there's nothing subtle going on here at all -- just definitions of various functions.
Best Answer
Hint: Recall that a geodesic is uniquely determined by its initial point and initial direction.