I hate to throw cold water on the party, but surprisingly, the formula that the OP was trying to prove ($[X',Y']=[X,Y]'$) is actually false when $G$ acts on the left. This formula is correct if $G$ acts on the right on $M$; but if $G$ acts on the left, then the correct formula is $[X',Y']=-[X,Y]'$.
Here's how to prove it. Suppose first that $G$ acts smoothly on $M$ on the right. Fix $p\in M$, and consider the orbit map $\alpha^{(p)}: G\to M$ defined by $\alpha^{(p)}(g) = p\cdot g$. Then I claim that for each $X\in \mathfrak g$, the fundamental vector field $X'$ is $\alpha^{(p)}$-related to $X$. To see this, note that the group law $p\cdot gg' = (p\cdot g)\cdot g'$ translates to $$\alpha^{(p)}\circ L_g (g') = \alpha^{(p\cdot g)}(g')$$
(where $L_g$ is left multiplication by $g$).
Let $g\in G$ be arbitrary and set $q=p\cdot g=\alpha^{(p)}(g)$. Because $X$ is left-invariant,
$$ X'_q = d\bigl(\alpha^{(q)}\bigr)_1(X_1) = d\bigl(\alpha^{(p)}\bigr)_g\circ d(L_g)_1(X_1) =
d\bigl(\alpha^{(p)}\bigr)_g(X_g)$$
(where the first equality is essentially the definition of $X'$, and the second follows from the previous equation by taking differentials at the identity).
This proves the claim.
Because brackets of $\alpha^{(p)}$-related vector fields are themselves $\alpha^{(p)}$-related, it follows that $[X',Y']_p = ([X,Y]')_p$, and since this is true for every $p$ it is true globally.
Now if $G$ acts on $M$ on the left, the above argument doesn't work; if $\tilde\alpha^{(p)}$ denotes the orbit map for the left action, then $X'$ is not $\tilde\alpha^{(p)}$-related to $X$. Instead, we can create a right action by setting $p\cdot g = g^{-1}\cdot p$. Letting $\alpha^{(p)}$ denote the orbit map for this new right action, we have $\tilde\alpha^{(p)}=\alpha^{(p)}\circ\iota$, where $\iota: G\to G$ is inversion; and the argument above shows that $X$ and $X'$ are $\alpha^{(p)}$-related. Since $d\alpha^{(p)}$ preserves brackets and $d\iota$ reverses them, it follows that $d\tilde\alpha^{(p)}$ reverses brackets.
The problem with Eric's argument has to do with the identification between $\operatorname{Lie}(\operatorname{Diff}(M))$ and $\mathfrak X(M)$. Essentially the same argument as above shows that the natural map $\operatorname{Lie}(\operatorname{Diff}(M)) \to \mathfrak X(M)$ is actually a Lie algebra anti-isomorphism (if we consider $\operatorname{Diff}(M)$ as an infinite-dimensional Lie group acting on $M$ on the left).
The problem with Victor Protsak's argument is that it doesn't actually show that $(X,0)$ (considered as a vector field on $G\times M$) is $\alpha$-related to $X'$ -- to show this, you'd have to prove that $(d\alpha)_{(g,m)}(X,0)=X'$ for every $g$ in the group, not just $g=1$.
I had to sort all this out recently because I'm adding a section on infinitesimal group actions in the second edition of my book Introduction to Smooth Manifolds. When I tried to prove the same formula the OP was trying to prove, I was surprised to find that it led to contradictions. Eventually I came up with the argument I sketched here, and then found at least one other book that confirms the result. I'm at home now and don't have access to my books, but if you'd like I'll find the reference and post it tomorrow.
Another approach.
To show it's impossible (the rank can't be 1), it is enough to show this when the field (assumed of char 0) is algebraically closed, and in turn it's enough to show the result in case $\mathfrak{g}$ is simple. If $x$ has $\mathrm{ad}(x)$ of rank 1, $x$ has centralizer of codimension 1. It is known (see e.g. this MathSE answer) that $\mathfrak{g}$ has no subalgebra of codimension 1, unless $\mathfrak{g}$ is 3-dimensional. But for $\mathfrak{sl}_2$, the operator $\mathrm{ad}_x$ has rank 2 for every nonzero $x$ (alternatively, all 2-dimensional subalgebras have a trivial centralizer, so can't be centralizer of an element).
A similar approach can be used (with a little further work) to classify which $\mathfrak{g}$ admit an $x$ with $\mathrm{ad}(x)$ of rank 2, and classify such elements $x$.
Best Answer
What about the formula $(D_X\varphi)(g)=\frac{\rm d}{{\rm d}t}\Bigl\vert_{t=0}\varphi(g\exp_G(tX))$ for $\varphi\in C^\infty(G)$, $g\in G$, and $X\in \mathfrak g$, where $G$ is a Lie group with its Lie algebra $\mathfrak g$? See for instance Eq. (5) in Ch. II of the book by S.Helgason, "Differential geometry, Lie groups, and symmetric spaces".