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Apparently, under Chebyshev transform of a generating function $A(x)$ OP understands a function $B(x) := C(-x^2)A(xC(-x^2))$, where $C(x):=\frac{1-\sqrt{1-4x}}{2x}$ is the generating function for Catalan numbers.

Let $m$ be fixed.

Consider the generating function $$\mathcal{R}(x,y):=\sum_{n=0}^\infty\sum_{j=0}^\infty R(n,j,m)x^jy^n$$ and so $\mathcal{R}(x,0) = (1-(m-1)x)^{-1}$.

Then the recurrence formula for $R(n,q,m)$ translates into $$(\star)\quad\mathcal{R}(x,y)-\mathcal{R}(x,0) = \frac{y}x(\mathcal{R}(x,y)-\mathcal{R}(0,y)) + \frac{y}{1+x}\mathcal{R}(x,y),$$ and we need to show that $\mathcal{R}(0,y) = \frac{C(-y^2)}{1-myC(-y^2)}$, which translates into $$\mathcal{R}(x,y) = \frac{(1+x)(x-(1+x)yC(-y^2))}{(x+x^2-(1+2x)y)(1-(m-1)x)(1-myC(-y^2))}.$$

Since the recurrence equation $(\star)$ together with the value $\mathcal{R}(x,0)$ uniquely determines $\mathcal{R}(x,y)$, it's enough just to verify that the purported formula for $\mathcal{R}(x,y)$ satisfies $(\star)$ as $x\to 0$, that is, $$\mathcal{R}(0,y)-1 = y\mathcal{R}'_x(0,y) + y\mathcal{R}(0,y),$$ and Sage verifies it.

UPDATED. The argument below is corrected.

Apparently, under Chebyshev transform of a generating function $A(x)$ OP understands a function $B(x) := C(-x^2)A(xC(-x^2))$, where $C(x):=\frac{1-\sqrt{1-4x}}{2x}$ is the generating function for Catalan numbers.

Let $m$ be fixed.

Consider the generating function $$\mathcal{R}(x,y):=\sum_{n=0}^\infty\sum_{j=0}^\infty R(n,j,m)x^jy^n$$ and so $\mathcal{R}(x,0) = f(x)$ is given.

Then the recurrence formula for $R(n,q,m)$ translates into $$(\star)\quad\mathcal{R}(x,y)-\mathcal{R}(x,0) = \frac{y}x(\mathcal{R}(x,y)-\mathcal{R}(0,y)) + \frac{y}{1+x}\mathcal{R}(x,y),$$ from where we need to deduce the value of $g(y) := \mathcal{R}(0,y)$. Expressing $\mathcal{R}(x,y)$ in the last equation, we get $$\mathcal{R}(x,y) = \frac{1+x}{x+x^2-(1+2x)y}\big(xf(x)-yg(y)\big).$$ Here the series $xf(x)-yg(y)$ must be divisible by the polynomial $x+x^2-(1+2x)y$. Plugging in its zero $\frac{\sqrt{1+4y^2}-1+2y}2=C(-y^2)y^2+y$, we get $$g(y) = (C(-y^2)y+1)f\big(C(-y^2)y^2+y\big).$$

Plugging $f(x) = (1-(m-1)x)^{-1}$ in the general solution, we get $$g(y) = (C(-y^2)y+1)\frac1{1-(m-1)(C(-y^2)y^2+y)}.$$ To show that it is the same as required $\frac{C(-y^2)}{1-myC(-y^2)}$, compute their difference: $$g(y) - \frac{C(-y^2)}{1-myC(-y^2)} = \frac{1-C(-y^2)-C(-y^2)^2y^2}{(1-(m-1)(C(-y^2)y^2+y))(1-myC(-y^2))},$$ which is zero since $1-C(-y^2)-C(-y^2)^2y^2=0$ as follows from the definition of $C$.

Apparently, under Chebyshev transform of a generating function $A(x)$ OP understands a function $B(x) := C(-x^2)A(xC(-x^2))$, where $C(x):=\frac{1-\sqrt{1-4x}}{2x}$ is the generating function for Catalan numbers.

Let $m$ be fixed.

Consider the generating function $$\mathcal{R}(x,y):=\sum_{n=0}^\infty\sum_{j=0}^\infty R(n,j,m)x^jy^n$$ and so $\mathcal{R}(x,0) = (1-(m-1)x)^{-1}$.

Then the recurrence formula for $R(n,q,m)$ translates into $$(\star)\quad\mathcal{R}(x,y)-\mathcal{R}(x,0) = \frac{y}x(\mathcal{R}(x,y)-\mathcal{R}(0,y)) + \frac{y}{1+x}\mathcal{R}(x,y),$$ and we need to show that $\mathcal{R}(0,y) = \frac{C(-y^2)}{1-myC(-y^2)}$, which translates into $$\mathcal{R}(x,y) = \frac{1+x}{x+x^2-(1+2x)y}\left(\frac{x}{1-(m-1)x} - \frac{yC(-y^2)}{1-myC(-y^2)}\right).$$

Since the recurrence equation $(\star)$ together with the value $\mathcal{R}(x,0)$ uniquely determines $\mathcal{R}(x,y)$, it's enough just to verify that the purported formula for $\mathcal{R}(x,y)$ satisfies $(\star)$ as $x\to 0$, that is, $$\mathcal{R}(0,y)-1 = y\mathcal{R}'_x(0,y) + y\mathcal{R}(0,y),$$ and Sage verifies it.

Apparently, under Chebyshev transform of a generating function $A(x)$ OP understands a function $B(x) := C(-x^2)A(xC(-x^2))$, where $C(x):=\frac{1-\sqrt{1-4x}}{2x}$ is the generating function for Catalan numbers.

Let $m$ be fixed.

Consider the generating function $$\mathcal{R}(x,y):=\sum_{n=0}^\infty\sum_{j=0}^\infty R(n,j,m)x^jy^n$$ and so $\mathcal{R}(x,0) = (1-(m-1)x)^{-1}$.

Then the recurrence formula for $R(n,q,m)$ translates into $$(\star)\quad\mathcal{R}(x,y)-\mathcal{R}(x,0) = \frac{y}x(\mathcal{R}(x,y)-\mathcal{R}(0,y)) + \frac{y}{1+x}\mathcal{R}(x,y),$$ and we need to show that $\mathcal{R}(0,y) = \frac{C(-y^2)}{1-myC(-y^2)}$, which translates into $$\mathcal{R}(x,y) = \frac{(1+x)(x-(1+x)yC(-y^2))}{(x+x^2-(1+2x)y)(1-(m-1)x)(1-myC(-y^2))}.$$

Since the recurrence equation $(\star)$ together with the value $\mathcal{R}(x,0)$ uniquely determines $\mathcal{R}(x,y)$, it's enough just to verify that the purported formula for $\mathcal{R}(x,y)$ satisfies $(\star)$ as $x\to 0$, that is, $$\mathcal{R}(0,y)-1 = y\mathcal{R}'_x(0,y) + y\mathcal{R}(0,y),$$ and Sage verifies it.

Apparently, under Chebyshev transform of a generating function $A(x)$ OP understands a function $B(x) := C(-x^2)A(xC(-x^2))$, where $C(x):=\frac{1-\sqrt{1-4x}}{2x}$ is the generating function for Catalan numbers.

Let $m$ be fixed.

Consider the generating function $$\mathcal{R}(x,y):=\sum_{n=0}^\infty\sum_{j=0}^\infty R(n,j,m)x^jy^n$$ and so $\mathcal{R}(x,0) = (1-(m-1)x)^{-1}$.

Then the recurrence formula for $R(n,q,m)$ translates into $$(\star)\quad\mathcal{R}(x,y)-\mathcal{R}(x,0) = \frac{y}x(\mathcal{R}(x,y)-\mathcal{R}(0,y)) + \frac{y}{1+x}\mathcal{R}(x,y),$$ and we need to show that $\mathcal{R}(0,y) = \frac{C(-y^2)}{1-myC(-y^2)}$, which translates into $$\mathcal{R}(x,y) = \frac{x(1+x)}{(1-(m-1)x)(x+x^2-(1+2x)y)} - \frac{(1+x)y}{x+x^2-(1+2x)y}\frac{C(-y^2)}{1-myC(-y^2)}.$$

Since the recurrence equation $(\star)$ together with the value $\mathcal{R}(x,0)$ uniquely determines $\mathcal{R}(x,y)$, it's enough just to verify the purported formula for $\mathcal{R}(x,y)$ satisfies $(\star)$ as $x\to 0$, that is, $$\mathcal{R}(0,y)-1 = y\mathcal{R}'_x(0,y) + y\mathcal{R}(0,y),$$ and Sage verifies it.

Apparently, under Chebyshev transform of a generating function $A(x)$ OP understands a function $B(x) := C(-x^2)A(xC(-x^2))$, where $C(x):=\frac{1-\sqrt{1-4x}}{2x}$ is the generating function for Catalan numbers.

Let $m$ be fixed.

Consider the generating function $$\mathcal{R}(x,y):=\sum_{n=0}^\infty\sum_{j=0}^\infty R(n,j,m)x^jy^n$$ and so $\mathcal{R}(x,0) = (1-(m-1)x)^{-1}$.

Then the recurrence formula for $R(n,q,m)$ translates into $$(\star)\quad\mathcal{R}(x,y)-\mathcal{R}(x,0) = \frac{y}x(\mathcal{R}(x,y)-\mathcal{R}(0,y)) + \frac{y}{1+x}\mathcal{R}(x,y),$$ and we need to show that $\mathcal{R}(0,y) = \frac{C(-y^2)}{1-myC(-y^2)}$, which translates into $$\mathcal{R}(x,y) = \frac{1+x}{x+x^2-(1+2x)y}\left(\frac{x}{1-(m-1)x} - \frac{yC(-y^2)}{1-myC(-y^2)}\right).$$

Since the recurrence equation $(\star)$ together with the value $\mathcal{R}(x,0)$ uniquely determines $\mathcal{R}(x,y)$, it's enough just to verify that the purported formula for $\mathcal{R}(x,y)$ satisfies $(\star)$ as $x\to 0$, that is, $$\mathcal{R}(0,y)-1 = y\mathcal{R}'_x(0,y) + y\mathcal{R}(0,y),$$ and Sage verifies it.