Vtyjkyuk

Some preliminary definitions. On page 342 of Lee's Smooth Manifold he concludes that $hatg(operatornamegrad f)(X) = Xf$ where $hatg$ is the isomorphism between $TM to T^*M$, the tangent bundle and co-tangent bundle.

Now according to Lee, $hatg^-1(omega) = g^ijomega_j fracpartial partial x^j$

But what I found is that,

beginalign hatg(hatg^-1(df))(X)
&=hatg(g^ijfracpartial fpartial x^j fracpartial partial
x^j)(X)\
&=g_ijg^ijfracpartial fpartial x^jfracpartial partial
x^j X^i\ &=fracpartial fpartial x^jfracpartial partial
x^j X^i \ &= X^i fracpartial fpartial x^j fracpartial
partial x^j textbecause g is symmetric, so we can bring X to the
front
endalign

But $Xf = X^i(x)fracpartial fpartial x^j$. So there is an extra basis term coming out of my reduction. What's going on?

First of all you have mixed the indices a bit. In the first line we should have

$$hatg(hatg^-1(df))(X)
=hatgleft(g^ijfracpartial fpartial x^jfracpartial partial
x^colorrediright) X$$

For the second line you shouldn't keep $Y=g^ijfracpartial fpartial x^jfracpartial partial x^i$ when you evaluate $g$. Instead you should just take the $k$-th component. This is explicitly stated at the beggining of p. 342. Indeed $hatg(Y)(X)$ should produce a number, not a vector like in your case. Thus we would get:

$$hatg(hatg^-1(df))(X)
=hatgleft(g^ijfracpartial fpartial x^jfracpartial partial
x^iright) X = g_ik X^i g^kjfracpartial fpartial x^j = delta^j_i X^i fracpartial fpartial x^j = X^ifracpartial fpartial x^i = Xf$$

Note that $Xf = X^ifracpartial fpartial x^colorredi$, not $Xf = X^ifracpartial fpartial x^j$, as you have mentioned in the last line.

REMARK: The reason why I take the $k$-th component is because we have already used $j$ for the coordinate representation of $hatg^-1(df)$. Thus $g(X)(Y) = g_ijX^iY^k$. In fact I believe that this is the reason behind your confusion in the probem.