How can I solve the discrete logarithm modulo 2q+1 if I can solve it in the subgroup of order q?
Ok, lets try again from scratch.  First, some group theory definitions:


A group is a set of elements and an operation $\times$ which satisfies a specific set of properties (closure, associativity, an identity element, inverses)
A subgroup is a subset of elements of a group which, along with the group operation from the main group, satisfies those same set of properties (closure, associativity, an identity element, inverses).  Note that the subgroup uses the exact same group operation $\times$, except that it restricts the operands to the subset.


For a finite group $G$, a subset of elements $H$ is a subgroup iff the subgroup is closed; that is, if you take any two elements $h_1, h_2$, then $h_1 \times h_2$ is also an element of $H$.

Now, $\mathbb Z^∗_p$ is a group with $p-1$ elements; the operation $a \times b$ is defined to be $ab \bmod p$; this satisfies all the group properties.

Now, consider only those elements $x$ of $\mathbb Z^∗_p$ that can be represented as $x = y^2 \mod p$ for some value of $y$ (or, in other words, that quadratic equation has at least one solution for $y$).  This turns out to be closed; that is, if we have two such elements $a = c^2$ and $b = d^2$, then $a \times b = ab = c^2d^2 = (cd)^2$ is also within this subset, and so these elements make up a subgroup.  It turns out that this subgroup has $(p-1)/2 = q$ elements (and is the only subgroup of that size), and hence is the "subgroup of order q" of the question.

With that background in place, the question under discussion is: suppose we had an oracle that could, given values $s$ and $t$ that were point a part of this subgroup (that is, there are some values $u$ and $v$ such that $s = u^2$, $t = v^2$) it could give us the value $x$ such that $t = s^x \bmod p$ (note: this is $\bmod p$, not $\bmod q$, because the subgroup shares the same operator with the main group).  Now, with such an oracle, how can we solve the DLP in the full group, that is, given $u$ and $v$, either of which may or may not be a member of the subgroup, how can we solve $v = u^x \bmod p$?

The first thing to note is if we define $s = u^2$ and $t = v^2$, both $s$ and $t$ are members of this subgroup.  Hence, we can use the oracle to find $y$ with $t = s^y \bmod p$.  Replacing the definitions for $s$ and $t$, we get $v^2 = u^{2y} \bmod p$.  Now, if there is a solution $v = u^x \bmod p$, we know that $v^2 = u^{2x \mod p-1}$ (Fermat's Little Theorem), we know that either $2y = 2x \bmod p-1$, and since $p-1$ is even, either $y = x$ or $y + (p-1)/2 = x$