The mass of the particle transforms from one coordinate to another.
You make reference to the old notion of the relativistic mass
\begin{equation}
m\boldsymbol{=}\dfrac{m_{\rm o}}{\sqrt{1\boldsymbol{-}\dfrac{v^2}{c^2}}}\boldsymbol{=}\gamma_v\,m_{\rm o}
\tag{01}\label{01}
\end{equation}
where $\,m_{\rm o}\,$ the rest mass. The relativistic mass $\,m\,$ is a source of confusion and it's not in use in contemporary Physics.
The rest mass $\,m_{\rm o}\,$ is a Lorentz invariant scalar.
What about the other quantity like force? Is there any four-vector for them? Is it, in general, I can find four-vector for euclidian vectors in Minkowski space?
From a rest mass $\,m_{\rm o}\,$ preserving 3-vector force $\,\mathbf f\,$ (like the Lorentz force of Electrodynamics) applied on a particle moving with 3-velocity $\,\mathbf u\,$ we produce a Lorentz 4-vector force $\,\mathbf F\,$ as follows
\begin{equation}
\mathbf F\boldsymbol{=}\left(\gamma_{\mathrm u}\mathbf{f}, \gamma_{\mathrm u}\dfrac{\mathbf{f}\boldsymbol{\cdot}\mathbf{u}}{c}\right)\,, \qquad \gamma_{\mathrm u}\boldsymbol{=}\left(1\boldsymbol{-}\dfrac{\mathrm u^2}{c^2}\right)^{\boldsymbol{-}\tfrac{1}{2}}
\tag{02}\label{02}
\end{equation}
Under a Lorentz boost with velocity $\,\boldsymbol{\upsilon}\,$ the rest mass $\,m_{\rm o}\,$ preserving 3-vector force $\,\mathbf f\,$ is transformed as follows
\begin{equation}
\mathbf f' = \dfrac{\mathbf f\boldsymbol{+}\dfrac{\gamma_v^2}{c^2 \left(\gamma_v\boldsymbol{+}1\right)}\left(\mathbf f\boldsymbol{\cdot} \boldsymbol{\upsilon}\right)\boldsymbol{\upsilon}-\gamma_v \boldsymbol{\upsilon}\left(\dfrac{\mathbf f\boldsymbol{\cdot}\mathbf u}{c^{2}}\right)}{\gamma_v \left(1-\dfrac{\boldsymbol{\upsilon}\boldsymbol{\cdot}\mathbf{u}}{c^{2}}\right)}
\tag{03}\label{03}
\end{equation}
Some details in my answer here Are magnetic fields just modified relativistic electric fields? would be useful.\
Not all 3-vectors, say $\,\mathbf h\,$, have a Lorentz 4-vector partner. But if so then you must $''$discover$''$ two scalars $\,\rm a_{\rm h},b_{\rm h}\,$ so that to build the Lorentz 4-vector
\begin{equation}
\mathbf H\boldsymbol{=}\left(\rm a_{\rm h}\,\mathbf h, \rm b_{\rm h}\right)
\tag{04}\label{04}
\end{equation}
For example in case of the rest mass $\,m_{\rm o}\,$ preserving 3-vector force $\,\mathbf f\,$ in equation \eqref{02}
\begin{equation}
\mathrm a_{\rm f}\boldsymbol{=}\gamma_{\mathrm u}\,,\qquad b_{\rm f}\boldsymbol{=}\gamma_{\mathrm u}\dfrac{\mathbf f\boldsymbol{\cdot}\mathbf u}{c}
\tag{05}\label{05}
\end{equation}
An easy and safe method to build new Lorentz 4-vectors is to take a Lorentz invariant scalar multiple of a known Lorentz 4-vector. By this method we build the Lorentz 4-vectors :
$\boldsymbol{\S\,1.} \texttt{ The Velocity 4-vector } \mathbf U$
The velocity 4-vector $\,\mathbf U\,$ is built by differentiation of the space-time position Lorentz 4-vector $\,\mathbf X\boldsymbol{=}\left(\mathbf x, c\,t\right)\,$ with respect to the Lorentz invariant scalar proper time $\,\tau$. Note that these properties follow the differentials $\,\mathrm d\mathbf X\,$ and $\,\mathrm d\tau\,$ respectively.
\begin{equation}
\!\!\!\!\!\!\!\!\!\!\!\!\!\!\mathbf U\boldsymbol{=}\dfrac{\mathrm d\mathbf X}{\mathrm d\tau}\boldsymbol{=}\left(\dfrac{\mathrm d\mathbf x}{\mathrm d\tau}, c\dfrac{\mathrm d t}{\mathrm d\tau}\right)\boldsymbol{=}\left(\dfrac{\mathrm d\mathbf x}{\mathrm dt}\dfrac{\mathrm d t}{\mathrm d\tau}, c\dfrac{\mathrm d t}{\mathrm d\tau}\right)\boldsymbol{=}\left( \gamma_{\mathrm u}\,\mathbf u, \gamma_{\mathrm u}\, c\right)\,, \:\: \gamma_{\mathrm u}\boldsymbol{=}\left(1\boldsymbol{-}\dfrac{\mathrm u^2}{c^2}\right)^{\boldsymbol{-}\tfrac{1}{2}}
\tag{06}\label{06}
\end{equation}
Under a Lorentz boost with velocity $\,\boldsymbol{\upsilon}\,$ the velocity 3-vector $\,\mathbf u\,$ is transformed as follows
\begin{equation}
\mathbf u' = \dfrac{\mathbf u\boldsymbol{+}\dfrac{\gamma_v^2}{c^2 \left(\gamma_v\boldsymbol{+}1\right)}\left(\mathbf u\boldsymbol{\cdot} \boldsymbol{\upsilon}\right)\boldsymbol{\upsilon}-\gamma_v \boldsymbol{\upsilon}}{\gamma_v \left(1-\dfrac{\boldsymbol{\upsilon}\boldsymbol{\cdot}\mathbf{u}}{c^{2}}\right)}
\tag{07}\label{07}
\end{equation}
$\boldsymbol{\S\,2.} \texttt{ The Linear Momentum 4-vector } \mathbf P$
The linear momentum 4-vector $\,\mathbf P\,$ is built as the Lorentz invariant rest mass $\,m_{\rm o}\,$ scalar multiple of the velocity Lorentz 4-vector $\,\mathbf U$.
\begin{equation}
\mathbf P\boldsymbol{=}m_{\rm o}\,\mathbf U\boldsymbol{=}\left( \gamma_{\mathrm u}\,m_{\rm o}\,\mathbf u, \gamma_{\mathrm u}\,m_{\rm o}\, c\right)\boldsymbol{=} \left( \mathbf p, \rm E/c\right)
\tag{08a}\label{08a}
\end{equation}
where
\begin{align}
\mathbf p & \boldsymbol{=}\gamma_{\mathrm u}\,m_{\rm o}\,\mathbf u\boldsymbol{=}\texttt{the linear momentum 3-vector}
\tag{08b}\label{08b}\\
\mathrm E & \boldsymbol{=}\gamma_{\mathrm u}\,m_{\rm o}\,c^2\boldsymbol{=}\texttt{ energy of the particle}
\tag{08c}\label{08c}
\end{align}
$\boldsymbol{\S\,3.} \texttt{ The Acceleration 4-vector } \mathbf A$
The acceleration 4-vector $\,\mathbf A\,$ is built by differentiation of velocity 4-vector $\,\mathbf U\,$ with respect to the Lorentz invariant scalar proper time $\,\tau$.
\begin{equation}
\mathbf A\boldsymbol{=}\dfrac{\mathrm d\mathbf U}{\mathrm d\tau}
\tag{09}\label{09}
\end{equation}
$\boldsymbol{\S\,4.} \texttt{ The Force 4-vector } \mathbf F$
For a rest mass $\,m_{\rm o}\,$ preserving 3-vector force $\,\mathbf f\,$ the force 4-vector $\,\mathbf F\,$ is built by differentiation of Linear Momentum 4-vector $\,\mathbf P\,$ with respect to the Lorentz invariant scalar proper time $\,\tau$.
\begin{equation}
\mathbf F\boldsymbol{=}\dfrac{\mathrm d\mathbf P}{\mathrm d\tau}\boldsymbol{=}\dfrac{\mathrm d\left(m_{\rm o}\,\mathbf U\right)}{\mathrm d\tau}\boldsymbol{=}m_{\rm o}\,\dfrac{\mathrm d\mathbf U}{\mathrm d\tau}\boldsymbol{=}m_{\rm o}\,\mathbf A
\tag{10}\label{10}
\end{equation}
