The answer is 'no' for $n=2$ (and hence for all higher $n$). Here is how one can see this.

First, when $n=2$, recall that, by the Gauss Lemma, a metric $g$ in geodesic normal coordinates $(x,y)$ centered on $p$ takes the form $$ g = \mathrm{d}x^2 + \mathrm{d}y^2 + h(x,y)\bigl(x\,\mathrm{d}y-y\,\mathrm{d}x)^2, $$ where the function $h$ is arbitrary, subject to the condition that $(x^2{+}y^2)h(x,y)+1>0$.

Letting $r^2 = x^2 + y^2$ and letting $R$ be the radial vector field $x\,\partial_x + y\,\partial_y$, one computes the formula for the Gauss curvature of $g$ to be $$ K = -\frac{2(1+r^2h)(RRh) - r^2(Rh)^2+4(3+r^2h)(Rh) + 8r^2h^2+12h}{4(1+r^2h)^2}. $$ Thus, in the geodesic disk of radius $\epsilon>0$ about $p$, i.e., where $r^2=x^2 + y^2 \le\epsilon^2$, we can keep $|K|$ as small as we like merely by imposing sufficiently small bounds on $h$, $Rh$ and $RRh$, i.e., $h$ and its first two radial derivatives. More precisely, for any $M>0$, there exists a $\delta>0$ such that, if $|h|$, $|Rh|$ and $|RRh|$ are bounded by $\delta$ when $r\le\epsilon$, then $|K|\le M$ when $r\le \epsilon$.

Let $\rho(r)$ be a smooth function that is identically zero near $r=0$ and $r=\epsilon$ and, say, positive, at $r=\epsilon/2$, but satisfies the condition that, for any constant $\lambda$ with $|\lambda|\le 1$, the function $h(x,y) = \lambda\rho(r)$ yields a $K$ that satisfies the bound $|K|\le 1$.

Let $f(\theta)$ be any $2\pi$-periodic smooth function bounded by $1$ and consider the smooth function $$ h(r\,\cos\theta,r\,\sin\theta) = \rho(r)f(\theta). $$ Then $h$ and its radial derivatives are bounded in such a way that the Gauss curvature $K$ for the corresponding metric $g$ will be bounded in absolute value by $1$, but the 'angular derivative' of $h$, i.e., $xh_y-yh_x = \rho(r)f'(\theta)$, need not be bounded. In particular, by choosing $f$ appropriately (bounded by $1$ but with very large first derivatives), we can be sure that the coefficients of $g$ in this coordinate system, i.e., $$ g_{11} = 1 + y^2\,h(x,y),\qquad g_{12} = -xy\,h(x,y),\qquad g_{22} = 1+x^2\,h(x,y), $$ while bounded themselves, will have some very large first derivatives when $r = \epsilon/2$. In particular, there is no constant $C>0$ that would bound the first derivatives of these quantities independent of the choice of $f$.

The answer is 'no' for $n=2$ (and hence for all higher $n$). Here is how one can see this.

First, when $n=2$, recall that, by the Gauss Lemma, a metric $g$ in geodesic normal coordinates $(x,y)$ centered on $p$ takes the form $$ g = \mathrm{d}x^2 + \mathrm{d}y^2 + h(x,y)\bigl(x\,\mathrm{d}y-y\,\mathrm{d}x)^2, $$ where the function $h$ is arbitrary, subject to the condition that $(x^2{+}y^2)h(x,y)+1>0$.

Letting $r^2 = x^2 + y^2$ and letting $R$ be the radial vector field $x\,\partial_x + y\,\partial_y$, one computes the formula for the Gauss curvature of $g$ to be $$ K = -\frac{2(1+r^2h)(RRh) - r^2(Rh)^2+2(5+3r^2h)(Rh) + 8r^2h^2+12h}{4(1+r^2h)^2}. $$ Thus, in the geodesic disk of radius $\epsilon>0$ about $p$, i.e., where $r^2=x^2 + y^2 \le\epsilon^2$, we can keep $|K|$ as small as we like merely by imposing sufficiently small bounds on $h$, $Rh$ and $RRh$, i.e., $h$ and its first two radial derivatives. More precisely, for any $M>0$, there exists a $\delta>0$ such that, if $|h|$, $|Rh|$ and $|RRh|$ are bounded by $\delta$ when $r\le\epsilon$, then $|K|\le M$ when $r\le \epsilon$.

Let $\rho(r)$ be a smooth function that is identically zero near $r=0$ and $r=\epsilon$ and, say, positive, at $r=\epsilon/2$, but satisfies the condition that, for any constant $\lambda$ with $|\lambda|\le 1$, the function $h(x,y) = \lambda\rho(r)$ yields a $K$ that satisfies the bound $|K|\le 1$.

Let $f(\theta)$ be any $2\pi$-periodic smooth function bounded by $1$ and consider the smooth function $$ h(r\,\cos\theta,r\,\sin\theta) = \rho(r)f(\theta). $$ Then $h$ and its radial derivatives are bounded in such a way that the Gauss curvature $K$ for the corresponding metric $g$ will be bounded in absolute value by $1$, but the 'angular derivative' of $h$, i.e., $xh_y-yh_x = \rho(r)f'(\theta)$, need not be bounded. In particular, by choosing $f$ appropriately (bounded by $1$ but with very large first derivatives), we can be sure that the coefficients of $g$ in this coordinate system, i.e., $$ g_{11} = 1 + y^2\,h(x,y),\qquad g_{12} = -xy\,h(x,y),\qquad g_{22} = 1+x^2\,h(x,y), $$ while bounded themselves, will have some very large first derivatives when $r = \epsilon/2$. In particular, there is no constant $C>0$ that would bound the first derivatives of these quantities independent of the choice of $f$.

The answer is 'no', already, even for $n=2$ (and hence for all higher $n$). Here is how one can see this: First, note that, when $n=2$, one can always write a metric $g$ in geodesic normal coordinates $(x,y)$ centered on $p$ in the form $$ g = \mathrm{d}x^2 + \mathrm{d}y^2 + h(x,y)\bigl(x\,\mathrm{d}y-y\,\mathrm{d}x)^2 $$ and the function $h$ is arbitrary, subject to the condition that $(x^2{+}y^2)h(x,y)+1>0$.

Now, let $r^2 = x^2 + y^2$ and let $R$ be the radial vector field $x\,\partial_x + y\,\partial_y$. Then one finds the formula for the Gauss curvature of $g$ to be $$ K = -\frac{2(1+r^2h)(RRh) - r^2(Rh)^2+4(3+r^2h)(Rh) + 8r^2h^2+12h}{4(1+r^2h)^2}. $$ Note that, in the geodesic disk of radius $\epsilon>0$ about $p$, i.e., where $x^2 + y^2 <\epsilon^2$, we can keep $K$ as small as we like merely by imposing sufficiently small bounds on $h$, $Rh$ and $RRh$, i.e., $h$ and its first two radial derivatives.

Now, let $\rho(r)$ be a smooth function that is identically zero near $r=0$ and $r=\epsilon$ and, say, positive, at $r=\epsilon/2$, but satisfies the condition that, for the function $h(x,y) = \rho\bigl((x^2+y^2)^{1/2}\bigr)$, the above formula for $K$ satisfies the bound $|K|\le \delta$ for some $\delta<1$. (This gives me a little room for fudging.) Now let $f(\theta)$ be any $2\pi$-periodic smooth function bounded by $1$ and consider the candidate function $$ h(r\,\cos\theta,r\,\sin\theta) = \rho(r)f(\theta). $$ Then $h$ and its radial derivatives are bounded in such a way that the Gauss curvature $K$ for the corresponding metric $g$ will be bounded in absolute value by $1$, but the 'angular derivatives' of $h$ need not be bounded at all. In particular, by choosing $f$ appropriately (bounded by $1$ but with very large first derivatives), we can be sure that the coefficients of $g$ in this coordinate system, i.e., $$ g_{11} = 1 + y^2\,h(x,y),\qquad g_{12} = -xy\,h(x,y),\qquad g_{22} = 1+x^2\,h(x,y) $$ while bounded themselves, will have some very large first derivatives when $r = \epsilon/2$. In particular, there is no constant $C>0$ that would bound the first derivatives of these quantities independent of the choice of $f$.

The answer is 'no' for $n=2$ (and hence for all higher $n$). Here is how one can see this.

First, when $n=2$, recall that, by the Gauss Lemma, a metric $g$ in geodesic normal coordinates $(x,y)$ centered on $p$ takes the form $$ g = \mathrm{d}x^2 + \mathrm{d}y^2 + h(x,y)\bigl(x\,\mathrm{d}y-y\,\mathrm{d}x)^2, $$ where the function $h$ is arbitrary, subject to the condition that $(x^2{+}y^2)h(x,y)+1>0$.

Letting $r^2 = x^2 + y^2$ and letting $R$ be the radial vector field $x\,\partial_x + y\,\partial_y$, one computes the formula for the Gauss curvature of $g$ to be $$ K = -\frac{2(1+r^2h)(RRh) - r^2(Rh)^2+4(3+r^2h)(Rh) + 8r^2h^2+12h}{4(1+r^2h)^2}. $$ Thus, in the geodesic disk of radius $\epsilon>0$ about $p$, i.e., where $r^2=x^2 + y^2 \le\epsilon^2$, we can keep $|K|$ as small as we like merely by imposing sufficiently small bounds on $h$, $Rh$ and $RRh$, i.e., $h$ and its first two radial derivatives. More precisely, for any $M>0$, there exists a $\delta>0$ such that, if $|h|$, $|Rh|$ and $|RRh|$ are bounded by $\delta$ when $r\le\epsilon$, then $|K|\le M$ when $r\le \epsilon$.

Let $\rho(r)$ be a smooth function that is identically zero near $r=0$ and $r=\epsilon$ and, say, positive, at $r=\epsilon/2$, but satisfies the condition that, for any constant $\lambda$ with $|\lambda|\le 1$, the function $h(x,y) = \lambda\rho(r)$ yields a $K$ that satisfies the bound $|K|\le 1$.

Let $f(\theta)$ be any $2\pi$-periodic smooth function bounded by $1$ and consider the smooth function $$ h(r\,\cos\theta,r\,\sin\theta) = \rho(r)f(\theta). $$ Then $h$ and its radial derivatives are bounded in such a way that the Gauss curvature $K$ for the corresponding metric $g$ will be bounded in absolute value by $1$, but the 'angular derivative' of $h$, i.e., $xh_y-yh_x = \rho(r)f'(\theta)$, need not be bounded. In particular, by choosing $f$ appropriately (bounded by $1$ but with very large first derivatives), we can be sure that the coefficients of $g$ in this coordinate system, i.e., $$ g_{11} = 1 + y^2\,h(x,y),\qquad g_{12} = -xy\,h(x,y),\qquad g_{22} = 1+x^2\,h(x,y), $$ while bounded themselves, will have some very large first derivatives when $r = \epsilon/2$. In particular, there is no constant $C>0$ that would bound the first derivatives of these quantities independent of the choice of $f$.