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.
The vector field $X=\dot \gamma$ is only defined along the geodesic $\gamma$, not on an open set of $M$, so it does not make sense to calculate the Lie derivative $\mathcal{L}_X (g(X,Y))$. In order to show that $g(X,Y)$ is constant along $\gamma$, we need to show that $\nabla_X g(X,Y)$ vanishes.
We get
$$
\begin{align*}
\nabla_X g(X,Y) = g(\nabla_X X, Y) + g(X, \nabla_X Y) = 0
\end{align*}
$$
In the first equality we used that $\nabla g = 0$.
The first term is zero since $\gamma$ is a geodesic. The second term vanishes because $Y$ is Killing and $g(X,X)$ is constant:
$$
\begin{align*}
0 = (\mathcal{L}_Y g)(X,X) &= \mathcal{L}_Y (g(X,X)) - 2g(\mathcal{L}_Y X, X) \\
&= 2 g([X,Y], X) = 2 g(\nabla_X Y-\nabla_Y X, X) \\
&= 2 g(\nabla_X Y, X)- Yg(X,X) = 2 g(\nabla_X Y,X).
\end{align*}
$$
Best Answer
Hint: You will first have to understand how the fact that $X$ is a Killing field is expressed in terms of the covariant derivative $\nabla X$. This is a standard result, it can be directly derived from exapanding the equation $0=(\mathcal L_Xg)(Y,Z)$. Having done that, you can express $\tfrac{d}{dt}g(\gamma'(\gamma(t)),X(\gamma(t)))$ in terms of covariant derivatives (along the curve) and use the equations satisfied by $\gamma'$ and by $X$ to conclude that this vanishes.