In[31]:= eq1 = (D1 b D[cs[r, t], r] + v0 cs[r, t] == jm/(1 + jm/k cs[r, t])) /. {r -> r0}

Out[31]= v0 cs[r0, t] + b D1 
\!\(\*SuperscriptBox[\(cs\), 
TagBox[
RowBox[{"(", 
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[r0, t] == jm/(1 + (jm cs[r0, t])/k)

In[32]:= eq2 = 
 eq1 /. {cs -> (Cs[#1, #2] (S/m) &)} /. {Cs -> (C1[#1/(Sqrt[
           D1 b/m]), #2/(b /m)] &)} /. {r0 -> R0 Sqrt[(b D1)/m], 
    t -> T b /m} // Simplify

Out[32]= (S v0 C1[R0, T])/m + Sqrt[(b D1)/m] S 
\!\(\*SuperscriptBox[\(C1\), 
TagBox[
RowBox[{"(", 
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[R0, T] == (jm k m)/(k m + jm S C1[R0, T])

In[33]:= eq3 = 
 eq2 /. {v0 -> Rv b D1/r0, 
     Sqrt[(b D1)/m] -> r0/R0} /. {b D1 -> (r0/R0)^2 m} // Simplify

Out[33]= (r0 S (Rv C1[R0, T] + R0 
\!\(\*SuperscriptBox[\(C1\), 
TagBox[
RowBox[{"(", 
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[R0, T]))/R0^2 == (jm k m)/(
 k m + jm S C1[R0, T])

In[34]:= eq4 = R0/(r0 S)*eq3[[1]] == R0/(r0 S)*eq3[[2]] // Expand

Out[34]= (Rv C1[R0, T])/R0 + 
\!\(\*SuperscriptBox[\(C1\), 
TagBox[
RowBox[{"(", 
RowBox[{"1", ",", "0"}], ")"}],
Derivative],
MultilineFunction->None]\)[R0, T] == (jm k m R0)/(
 r0 S (k m + jm S C1[R0, T]))

In order to get the result of eq. 4, I used more than 4 steps of replacements. Would you like to help me with easier way? Thanks.